Class 11 Engineering Maths - Trigonometric Functions

If $\displaystyle \alpha , \beta , \gamma \epsilon \left ( 0, \frac{\pi }{2} \right )$ , then $\displaystyle\frac{\sin \left ( \alpha +\beta +\gamma \right )}{\sin \alpha +\sin \beta +\sin \gamma }< a$

Find a

$\sin \left ( \alpha +\beta +\gamma \right )=\sin \alpha \cos \beta \cos \alpha +\cos \alpha \sin \beta \cos \gamma +\cos \alpha \cos \beta \sin \gamma -\sin \alpha \sin \beta \sin \gamma$
or

$\sin \left ( \alpha +\beta +\gamma \right )-\sin \alpha -\sin \beta -\sin \gamma =\sin \alpha \left ( \cos \beta \cos \gamma -1 \right )+\sin \beta \left ( \cos \gamma \cos \gamma -1 \right )+\sin \gamma \left ( \cos \alpha \cos \beta -1 \right )-\sin \alpha \sin \beta \sin \gamma$
or

$\sin \left ( \alpha +\beta +\gamma \right )-\sin \alpha -\sin \beta -\sin \gamma < 0$
or

$\sin \left ( \alpha +\beta +\gamma \right )< \sin \alpha +\sin \beta +\sin \gamma$
or

$\displaystyle \frac{\sin \left ( \alpha +\beta +\gamma \right )}{\sin \alpha +\sin \beta +\sin \gamma }< 1$

Ans: 1

For any $\theta$ , state the value of:
$\sin ^{2}\theta+\cos ^{2} \theta$

$\sin^2 \theta + \cos^2 \theta$
=

$(\dfrac{P}{H})^2 + (\dfrac{B}{H})^2$ where P is Perpendicular, B is Base and H is Hypotenuse
=

$\dfrac{P^2 + B^2}{H^2}$
=

$\dfrac{H^2}{H^2}$ (using Pythagoras Theorem

$H^2 = P^2 + B^2$ )
=

$1$

Prove that $\displaystyle \frac{\cos 10^{0}+\sin 10^{0}}{\cos 10^{0}-\sin 10^{0}}=\tan a^{0}$

Find a

$\displaystyle \frac{\cos 10^{0}+\sin 10^{0}}{\cos 10^{0}-\sin 10^{0}}=\frac{1+\tan 10^{0}}{1-\tan 10^{0}}=\frac{\tan 45^{0}+\tan 10^{0}}{1-\tan 45^{0}\tan 10^{0}}=\tan \left ( 45^{0}10^{0} \right )=\tan 55^{0}$ (dividing by

$\cos 10^{0}$ )

Ans: 55

Prove that:
$\displaystyle cot^2\frac{\pi}{6}+cosec\frac{5\pi}{6}+3tan^2\frac{\pi}{6}=6$

$LHS=cot^{ 2 }\dfrac { \pi }{ 6 } +cosec\dfrac { 5\pi }{ 6 } +3tan^{ 2 }\dfrac { \pi }{ 6 }$

$={ \sqrt { 3 } }^{ 2 }+cosec\left( \pi -\dfrac { \pi }{ 6 } \right) +3\dfrac { 1 }{ { \sqrt { 3 } }^{ 2 } }$

$=4+cosec\dfrac { \pi }{ 6 }$

$=4+2=6$

Hence proved.

What is the value of $\displaystyle \left ( \tan A+\cot A \right )\sin A\times \cos A$ ?

$\displaystyle \left ( \tan A+\cot A \right )\sin A\times \cos A$

$\displaystyle =\left ( \frac{\sin A}{\cos A}+\frac{\cos A}{\sin A} \right )\times \sin A\times \cos A=1$

What is the value of $\displaystyle \sin ^{6 }\theta +\cos ^{6}\theta +3\sin ^{2}\theta \cos ^{2}\theta$ ?

$\displaystyle \because \sin ^{2}\theta +\cos ^{2}\theta =1$

$\displaystyle \therefore \left ( \sin ^{2}\theta +\cos ^{2}\theta \right )^{3}=1^{3}=1$

$\displaystyle \Rightarrow \sin ^{6}\theta +\cos ^{6}\theta +3\sin ^{2}\theta \cos ^{2}\theta \left ( \sin ^{2}\theta +\cos ^{2}\theta \right )$

$\displaystyle =1$

$\displaystyle \Rightarrow \sin ^{6}\theta +\cos ^{6}\theta +3\sin ^{2}\theta \cos ^{2}\theta =1$

What is the value of $\displaystyle \sin ^{2}25^{0}+\sin ^{2}65^{0}$ ?

$\displaystyle \sin 65^{0}=\sin \left ( 90^{0}-25^{0} \right )=\cos 25^{0}$

$\displaystyle \therefore \sin ^{2}25^{0}+\sin ^{2}65^{0}$

$\displaystyle =\sin ^{2}25^{0}+\cos ^{2}25^{0}$

$\displaystyle =1\left ( \because \sin ^{2}\theta +\cos ^{2}\theta =1 \right )$

Prove that:
$\displaystyle \sin^2\frac{\pi}{6}+\cos^2\frac{\pi}{3}-\tan^2\frac{\pi}{4}=-\frac{1}{2}$

$\displaystyle L.H.S=\sin^{ 2 }\frac { \pi }{ 6 } +\cos^{ 2 }\frac { \pi }{ 3 } -\tan^{ 2 }\frac { \pi }{ 4 } ={ \left( \frac { 1 }{ 2 } \right) }^{ 2 }+{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }-{ 1 }^{ 2 }=\frac { -1 }{ 2 }$

Therefore, L.H.S

$=$ R.H.S

Hence proved.

If $\displaystyle \cos \theta =\frac{2t}{1+t^{2}}$ , then find the value of $\displaystyle \tan \theta$ .

$\displaystyle \sin \theta =\sqrt{1-\cos ^{2}\theta }=\sqrt{1-\left ( \frac{2t}{1+t^{2}} \right )^{2}}$

$\displaystyle \Rightarrow \sqrt{\frac{\left ( 1+t^{2} \right )-4t^{2}}{\left ( 1+t^{2} \right )^{2}}}=\sqrt{\frac{\left ( 1-t^{2} \right )^{2}}{1+t^{2}}}=\frac{1-t^{2}}{1+t^{2}}$

$\displaystyle \therefore \tan \theta =\frac{1-t^{2}}{1+t}+\frac{2t}{1+t^{2}}=\frac{1-t^{2}}{2t}$

Write the principal value of $\cos^{-1}[\cos (680^o)]$ .

$\cos^{-1}[\cos (680^{o})]$

$=\cos ^{-1}[\cos (2 \times 360-40)]$

$=\cos^{-1}[\cos 40^o]$

$=40^o$

Prove that:

$\dfrac{sin\theta - 2sin^{3}\theta}{2cos^{3}\theta-cos\theta} =tan\theta$

LHS

$= \dfrac{sin\,\theta -2\,sin^{3}\theta }{2\,cos^{3}\theta -cos\theta }$

$= \dfrac{sin\,\theta (1-2\,sin^{2}\theta )}{cos\theta (2cos^{2}\theta -1)}$

$= \dfrac{sin\,\theta }{cos\,\theta }\left [ \dfrac{1-2sin^{2}\theta }{2(1-sin^{2}\theta )-1} \right ]$

$= tan\theta \left [ \dfrac{1-2sin^{2}\theta }{2-2sin^{2}\theta -1} \right ]$

$= tan\theta \left [ \dfrac{1-2sin^{2}\theta }{1-2sin^{2}\theta } \right ]$

$= tan\theta = RHS$

1112941_874508_ans_1cb80b9ca345474a961a50702814bf1f.jpeg

Find the value of $\cos^{2} 12^{\circ} + \cos^{2} 78^{\circ}$

$\cos^{2} 12^{o}+\cos^{2} 78^{o}$

$=\cos^{2}(90^{o}-78^{o})+\cos^{2}78^{o}$

$=\sin^{2} 78^{o}+\cos^{2} 78^{o}$ .....

$[\because \sin\theta=\cos(90^{o}-\theta)]$

$=1$ ......

$[\because \sin^{2}\theta + \cos^{2}\theta=1]$

Show that $\left( 1+\cot ^{ 2 }{ \theta } \right) \left( 1-\cos { \theta } \right) \left( 1+\cos { \theta } \right) =1\quad$ .

$\displaystyle (1+cot^2\theta)(1-cos\theta)(1+cos\theta)=(1+\frac{cos^2\theta}{sin^2\theta})(1-cos^2\theta)$

$\displaystyle =(\frac{sin^2\theta+cos^2\theta}{sin^2\theta})(sin^2\theta)$

$\displaystyle =sin^2\theta+cos^2\theta$

$\displaystyle =1$

Prove $\text{cosec}^{6}A - \cos^{6}A = 3\cot^{2}A \text{cosec}^{2} A + 1$ .

$\cos e{c^6}A - {\cot ^6}A = 3{\cot ^2}A\cos e{c^2}A + 1$

$\cos e{c^2}\theta - {\cot ^2}\theta = 1$

$3 \times 1 = 3\dfrac{1}{{{{\sec }^2}(90^\circ - A)}} \times {\sec ^2}(90^\circ - A) + 1$

$3 = \dfrac{1}{{{{\sec }^2}(90^\circ - A)}} \times {\sec ^2}(90^\circ - A) + 1$

$3 = 3 \times 1$

$3 = 3$ Proved

Simplify the following expression:
$\displaystyle 3 \, \cos \, x \, - \, 4 \, \sin \, x \,\cos \, x \, - \, \sin^2 \, x \, - \, 1$

$3\cos x-4\sin x\cos x-{\sin}^{2}x-1$

$=(3\cos x-1)-\sin x(4\cos x-\sin x)$

Solve $\sin\theta +\sin 5\theta =\sin 3\theta; 0\leq \theta \leq \pi$ .

On simplification,

$\sin 3\theta (2\cos 2\theta +1)=0$

$\therefore \theta =n\pi /3, \theta =n\pi \pm \pi/3$ for given range

$\theta =0, \pi/3$

$\therefore \theta =0, \pi/3$ .

Solve $\sin x+\sqrt{3}\cos x \geq 1$ .

Proceeding as in (a), we get

$\cos \left(x-\dfrac{\pi}{6}\right)\geq \dfrac{1}{2}=\cos\dfrac{\pi}{3}$

$\therefore -\pi/3\leq x-(\pi/6)\leq \pi/3$
or

$-\pi/6\leq x\leq \pi/2$ .

Prove the following identities :
$\dfrac{1 \, - \, \cos \, \theta}{1 \, + \,\cos \, \theta} \, = \, (cosec \, \theta \, - \, \cot \, \theta)^2$

L.H.S. = $\dfrac{1 \, - \, \cos \, \theta}{1 \, + \, \cos \, \theta}$

$\Rightarrow \, \,$ L.H.S. = $\dfrac{1 \, - \, \cos \, \theta}{1 \, + \, \cos \, \theta} \, \times \, \dfrac{1 \, - \, \cos \, \theta}{1 \, - \, \cos \, \theta}$

$\Rightarrow \, \,$ L.H.S. = $\dfrac{(1 \, - \, \cos \, \theta)^2}{1 \, - \, \cos ^2 \, \theta} \, = \, \dfrac{(1 \, - \, \cos \, \theta)^2}{sin^2 \, \theta}$

$\Rightarrow \, \,$ L.H.S. = $\left(\dfrac{1 \, - \, \cos \, \theta}{sin \, \theta}\right)^2 \, = \, \left(\dfrac{1}{\sin \, \theta} \, - \, \dfrac{\cos \, \theta}{\sin \, \theta}\right)^2 \, = \, (cosec \, \theta \, - \cot \, \theta)^2 \, = \, R.H.S.$

Solve $1-2\sin \theta -2\cos \theta +\cot \theta =0$ , $(0 < \theta < 2\pi)$ .

$(\cos\theta +\sin\theta)\left[\dfrac{1}{\sin\theta} -2\right]=0$

$\therefore \sin\theta =\dfrac{1}{2}, \tan\theta =-1=\tan\left(-\dfrac{\pi}{4}\right)$

$\therefore \theta =\dfrac{\pi}{6}, \pi -\dfrac{\pi}{6}, \pi+\left(-\dfrac{\pi}{4}\right), 2\pi +\left(-\dfrac{\pi}{4}\right)$

$\therefore \theta =\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{4}, \dfrac{7\pi}{4}$ in the given range.

Solve: $2+7\tan^2x=3.25sec^2x$ .

Ans.

$n\pi\pm \dfrac{\pi}{6}$ . Put

$sec^2x=1+\tan^2x$

and we get

$\dfrac{15}{4}t^2=\dfrac{5}{4}$

$\tan x=\pm\dfrac{1}{\sqrt{3}}=\pm \tan\dfrac{\pi}{6}$ etc.

Convert $230^\circ$ into radian.

1237666_1000452_ans_49965cc6cf294800868106b1145a3131.jpeg

Solve: $4\sin^4x+\cos^4x=1$ .

The given equation can be put in the form

$4\sin^4x=1-\cos^4x=(1-\cos^2x)(1+\cos^2x)$

$\sin^2x[4\sin^2x-1-(1-\sin^2x)]=0$

$\sin^2x(5\sin^2x-2)=0$

$\therefore \sin x=0$ or

$\sin x=\pm \sqrt{2/5}$

$x=n\pi$ or

$x=n\pi\pm\alpha$ ,

where

$\sin\alpha =\sqrt{2/5}$ .

Convert $\dfrac{3\pi}{7}$ in degrees.

$\left( \dfrac { 3\pi }{ 7 } \right) ={ \left( \dfrac { 3\pi }{ 7 } \times \dfrac { 180 }{ \pi } \right) }^{ 0 }={ \left( \dfrac { 540 }{ 7 } \right) }^{ 0 }$

$2^{\cos 2x}=3.2^{\cos^2x}-4$

$2^{2c^2-1}=3.2^{c^2}-4$ . Put

$2^{c^2}=y$
or

$y^2=2(3y-4)$
or

$y^2-6y+8=0$

$\therefore (y-4)(y-2)=0$

$\therefore y=2c^2=2^2$ or

$2^1$

$\therefore \cos^2x=2$ or

$\cos^2x=1$

$\therefore \sin^2x=0$ or

$x=n\pi$ .

The expression $\dfrac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}}$ reduces to:

$\cfrac { \tan { A } +\sec { A-1 } }{ \tan { A } -\sec { A+1 } } \\$

since

$\sec^2\theta=1+\tan^2\theta$

$\cfrac { \tan { A } +\sec { A}-1 -( \sec^2A-\tan^2A) }{ \tan { A } -\sec { A }+1 } \\$

Since

$A^2-B^2=(A+B)(A-B)$

$\tan {A} +\sec {A}$

Prove that:
$\cot (\pi -\theta) = -\cot \theta$

$(\pi - \theta)$ lies in second quadrant and

$\cot$ is negative in second quadrant

$\Rightarrow \cot (\pi-\theta) =-\cot \theta$

Prove that:
$\sec(\pi - \theta) = - \sec\theta$

$(\pi - \theta)$ lies in second quadrant and

$\sec$ is negative in the second quadrant

$\Rightarrow \sec (\pi-\theta) =-\sec \theta$

Prove that:
$\text{cosec} (\pi -\theta)= \text{cosec }\theta$

$(\pi - \theta)$ lies in the second quadrant and

$\text{cosec}$ is positive in the second quadrant

$\Rightarrow \text{cosec} (\pi-\theta) =\text{cosec }\theta$

Find the principal solution or solution of $\sin x=\dfrac {1}{\sqrt {2}}$ .

We know that

$\sin x=\dfrac { 1 }{ \sqrt { 2 } } =\sin { \dfrac { \pi }{ 4 } }$ and by allied angles, we also know that

$\sin (\pi-x)=\sin x$

$\therefore \quad \sin \dfrac {\pi}{4}=\sin \left(\pi-\dfrac {\pi}{4}\right)=\sin \dfrac {3\pi}{4}$

$\therefore \quad \sin \dfrac {\pi}{4}=\sin {3\pi}{4}=\dfrac {1}{\sqrt {2}}$
such that

$0<\dfrac {\pi}{4}<2\pi$ and

$0<\dfrac {3\pi}{4}<2\pi$

$\therefore$ the required principal solution are

$x=\dfrac {\pi}{4}$ and

$x=\dfrac {3\pi}{4}$

Prove that :
$\dfrac{1}{\sec \, \theta \, - \, \tan \, \theta} \, = \, \sec \, \theta \, + \, \tan \, \theta$

We have,
LHS = $\dfrac{1}{\sec \, \theta \, - \, \tan \, \theta}$

$\Rightarrow$ LHS = $\dfrac{1}{\sec \, \theta \, - \, \tan \, \theta} \, \times \, \dfrac{\sec \, \theta \, + \, \tan \, \theta}{\sec \, \theta \, + \, \tan \, \theta}$

$\Rightarrow$ LHS = $\dfrac{\sec \, \theta \, + \, \tan \, \theta}{\sec ^2 \, \theta \, - \, \tan ^2 \, \theta} \, = \, \dfrac{\sec \, \theta \, + \, \tan \, \theta}{1}$ $[\because \, \sec ^2 \, \theta \, - \, \tan ^2 \, \theta \, = \, 1]$

$\Rightarrow$ LHS = $\sec \, \theta \, + \, \tan \, \theta \, = \, RHS$

show that $\sqrt { 2+\sqrt { 2+\sqrt { 2+2\cos { 8\theta } } } } =2\cos { \theta }$ , $0<\theta<\dfrac{\pi}{8}$

According to the problem:

$\sqrt {2 + \sqrt {2 + \sqrt {2 + 2\cos 8\theta } } }$

implies that,

$= \sqrt {2 + \sqrt {2 + \sqrt {2 + \left( {1 + \cos 8\theta } \right)} } }$

implies that,

$= \sqrt {2 + \sqrt {2 + \sqrt {\left( {4{{\cos }^2}4\theta } \right)} } }$

implies that,

$= \sqrt {2 + \sqrt {2 + 2\cos 4\theta } }$

implies that,

$= \sqrt {2 + \left( {2\sqrt {1 + \cos 4\theta } } \right)}$

implies that,

$= \sqrt {\left( {2 + \sqrt {4{{\cos }^2}2\theta } } \right)}$

implies that,

$= \sqrt {2\left( {1 + \cos 2\theta } \right)}$

implies that,

$= \sqrt {4{{\cos }^2}\theta }$

Hence the result will be,

$= 2\cos \theta$

If $\theta$ is an acute angle and $sin \theta=\cos \theta$ find $3tan\theta$

$\theta$ is an acute angle

$\Rightarrow \theta < {90^ \circ }$

And $\sin \theta = \cos \theta$ if $\theta = {45^ \circ }$

$3\tan \theta$

$= 3\tan {45^ \circ }$

$= 3 \cdot 1$

$= 3$

Find principal solution for $\tan x =-1$ . $x\epsilon(\dfrac{\pi}{2} ,\pi)$

$\tan x=-1$

Here, tan is negative.

We know, tan is negative in 2nd and 4th quadrant.

Here,

$\theta = 45^{o}$

Value in 2nd quadrant =

$180^{o}-45^{o}=135^{o}$

Value in 4th quadrant =

$360^{o}-45^{o}=315^{o}$

So, principal solutions are

$x=135^{o}=\dfrac{3\pi}{4}$ and

$x=315^{o}=\dfrac{7\pi}{4}$ .

Convert ${290}^{o}$ into radian measure

$290^{o}=290\times \dfrac{\pi}{180}=\dfrac{29\pi}{18}$

Hence the answer is

$\dfrac{29\pi}{18}$

If $a\sin \left( {\theta + \alpha } \right) = b\sin \left( {\theta + \alpha } \right)$ show that $\cot \theta = \dfrac{{a\cos \alpha - b\cos \alpha }}{{b\sin \alpha - a\sin \alpha }}$ .

from the above equation

$asin(\theta + \alpha)=bsin(\theta + \alpha)$

$asin\theta cos\alpha+asin\alpha cos\theta =bsin\theta cos\alpha+bsin\theta cos\alpha$

$\dfrac{tan\theta}{tan\alpha}=\dfrac{b-a}{a-b}$

$tan\theta=\dfrac{sin\alpha (b-a)}{cos\alpha (a-b)}$

$cot\theta=\dfrac{acos\alpha -bcos\alpha}{bsin\alpha -asin\alpha}$

Find the $particular solution$ of $\tan x + \tan 2x + \tan x.\tan 2x = 1$ .

$\tan x + \tan 2x + \tan x.\tan 2x = 1$

$\tan x + \tan 2x = 1 - \tan x.\tan 2x$

$\dfrac{{\tan x + \tan 2x}}{{1 - \tan x.\tan 2x}} = 1$

$\tan \left( {x + 2x} \right) = 1$

$\tan 3x = \tan \left( {{\pi }} \right)$

$3x = n\pi + {\pi }$

$x = \dfrac{{n\pi }}{3} + \dfrac{\pi }{{3}}$

${\cos ^4}\theta + {\cos ^2}\theta = 1$ then show that ${\sec ^4}\theta - {\sec ^2}\theta = 1$

We have,

${\cos ^4}\theta + {\cos ^2}\theta = 1$

$\dfrac{1}{\sec ^4\theta }+\dfrac{1} {\sec ^2\theta } = 1$

$\dfrac{1+\sec^2\theta}{\sec ^4\theta } = 1$

$1+\sec^2\theta=\sec ^4\theta$

$\sec^4\theta-\sec ^2\theta =1$

Hence, proved.

Prove that $\sum a(\sin \,B - \sin C) = 0$

$a(sinB-sinC)+a(sinC-sinA)+a(sinA-sinB)$

$asinB-asinC+asinC-asinA+asinA-asinB$

$=0$

Solve equation for general solution $2(\sin \, x)^2 + (\sin \, 2x)^2 = 2$ .

$2\sin ^{2}x+(\sin 2x)^2=2$

$2\sin ^{2}x+(2\sin x \cos x)^2=2$

$2\sin ^{2}x+4\sin^2 x (1-\sin^2 x)=2$

$2\sin ^{2}x+4\sin^2 x -4 \sin ^4x=2$

$4 \sin ^4x-6\sin^2 x+2=0$

Solving the above quadratic equation, we get,

$\sin ^{2}x=\dfrac{1}{2}$ and

$\sin ^{2}x=1$

$\sin x=\pm \dfrac{1}{2}$ and

$\sin x=\pm 1$

So,

$x=(2n+1)\dfrac{\pi}{2}$ and

$x=(2n+1)\dfrac{\pi}{2}$

Find radian measure corresponding to the degree measure $- {37^ \circ }30'$

$\because 1°=60' \Rightarrow30'=\left(30\times\dfrac{ 1}{60}\right)^o$

$=0.5^o$

So,

$37^o 30'=37^o+0.5^o=37.5^o$

$\because$

$1^o=\dfrac{ \pi}{180}$ radian

$\therefore$

$37.5^o= 37.5\times\dfrac{\pi}{180}$ radian

$=\dfrac{37.5}{180}\pi$ radian

$=\dfrac{5}{24} \pi$ radian.

If $\sin { A } =\cfrac { 5 }{ 13 }$ then evaluate $\cos { A }$ and $\tan { A }$ .

$\displaystyle \sin A=\frac{5}{13};\cos A =\sqrt{1-\sin^{2} A}=\pm\frac{12}{13};\tan A=\frac{\displaystyle \frac{5}{13}}{\displaystyle\pm\frac{12}{13}}=\pm\frac{5}{12}$

If $\sin 9x = \sin x$ . Find the values of $x$ .

$\begin{array}{l}\sin 9x - \sin x = 0\\\sin C - \sin D = 2\sin \left( {\frac{{C - D}}{2}} \right)\cos \left( {\frac{{C + D}}{2}} \right)\\\sin 9x - \sin x = 2\sin \left( {\frac{{9x - x}}{2}} \right)\cos \left( {\frac{{9x + x}}{2}} \right)\\ = 2\sin \left( {4x} \right)\cos \left( {5x} \right) = 0\\\sin 4x = 0\\4x = n\pi ;n \in I\\x = \frac{{n\pi }}{4};n \in I\\\cos 5x = 0\\5x = (2n + 1)\frac{\pi }{2};n \in I\\x = (2n + 1)\frac{\pi }{{10}};n \in I\end{array}$

Solve : $\dfrac{1}{\text{cosec} \, \theta + \cot \, \theta} - \dfrac{1}{\sin \, \theta } = \dfrac{1}{\sin \, \theta} - \dfrac{1}{\text{cosec} \theta - \cot \theta}$

$\dfrac{1}{\csc \theta +\cot \theta }-\dfrac{1}{\sin \theta }=\dfrac{1}{\sin \theta }-\dfrac{1}{\csc \theta -\cot \theta }$

$\dfrac{1}{\dfrac{1}{\sin \theta } +\dfrac{\cos \theta }{\sin \theta } }-\dfrac{1}{\sin \theta }=\dfrac{1}{\sin \theta }-\dfrac{1}{\dfrac{1}{\sin \theta } -\dfrac{\cos \theta }{\sin \theta }}$

$\dfrac{\sin \theta }{1+\cos \theta }-\dfrac{1}{\sin \theta }=\dfrac{1}{\sin \theta }-\dfrac{\sin \theta }{1-\cos \theta }$

$\dfrac{\sin \theta }{1+\cos \theta }+\dfrac{\sin \theta }{1-\cos \theta }=\dfrac{1}{\sin \theta }+\dfrac{1}{\sin \theta }$

$2\dfrac{\sin \theta }{1-\cos^2 \theta }=2\dfrac{1}{\sin \theta }$

$\dfrac{\sin \theta }{\sin ^2\theta }=\dfrac{1}{\sin \theta }$

$\dfrac{1}{\sin \theta }=\dfrac{1}{\sin \theta }$

$1=1$

Solve $\sqrt { \cfrac { 1+\cos { 30 } }{ 1-\cos { 30 } } }$

$\displaystyle \sqrt{\frac{1+\cos 30}{1-\cos 30}}=\sqrt{\frac{(2+\sqrt{3})^{2}}{4-3}}=2+\sqrt{3}$

Prove $\sin ^2\theta +\cos ^2\theta =1$

$\sin \theta =\dfrac {\text{Opposite side to }\theta }{\text{Hypotenuse }}$

$\cos \theta =\dfrac {\text{Adjacent side to }\theta }{\text{Hypotenuse }}$

$\sin^2 \theta =\dfrac {(\text{Opposite side to }\theta)^2 }{(\text{Hypotenuse )}^2}$

$\cos ^2\theta =\dfrac {\text{(Adjacent side to }\theta)^2 }{\text{(Hypotenuse )}^2}$

$\sin ^2\theta +\cos ^2\theta$

$\dfrac {(\text{Opposite side to }\theta)^2 }{(\text{Hypotenuse )}^2}+$

$\dfrac {\text{(Adjacent side to }\theta)^2 }{\text{(Hypotenuse )}^2}$

$\dfrac{\text{Hypotenuse }^2}{\text{Htpotenuse}^2}=1$

Prove if $sinA = sinB$ then $\angle A = \angle B$ .

Given:

$\sin A = \sin B$

We know that of $y = \sin x$ , then:

$x = {\sin ^{ - 1}}y$

Also, we know that for $\theta$ ,

${\sin ^{ - 1}}\left( {\sin \theta } \right) = \theta$

Then, $\sin A = \sin B$

Implies: ${\sin ^{ - 1}}\left( {\sin A} \right) = {\sin ^{ - 1}}\left( {\sin B} \right)$

$\Rightarrow A = B$

If $\sin A = \sin B$ , then $\angle A = \angle B$

If $\tan { 2\theta =\cot { \left( \theta +{ 6 } \right) } }$ , where $2\theta$ and $\theta +6$ are acute angles, find the value of $\theta$ .

Given

$2\theta,\theta+6$ are acute angles

$\tan 2\theta=\cot (\theta+6)$

$\tan 2\theta=\tan (90-\theta-6)$

$\implies 2\theta=84-\theta$

$\implies 3\theta=84$

$\implies \theta=28$

Solve $\tan x + \cot x$

Solve :-

$\tan x+\cot x$

$=\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}$

$=\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\sin x\cos x}$

$=\dfrac{1}{\sin x\cos x}$

$=\sec x\tan x$

Hence, this is the answer.

prove that:

$1+\cot^{2} \theta = \csc^{2} \theta$

$\displaystyle 1+cot^{2}\theta = cosec^{2}\theta$

L.H.S

$\displaystyle 1+cot^{2}\theta = 1+\frac{cos^{2}\theta }{sin^{2}\theta }$

$\displaystyle [\because cot\theta = \frac{cos\theta }{sin\theta }] = \frac{sin^{2}\theta +cos^{2}\theta }{sin^{2}\theta }$

$\displaystyle [\because sin^{2}\theta +cos^{2}\theta = 1] = \frac{1}{sin^{2}\theta }$

$\displaystyle [\because \frac{1}{sin\theta = cosec\theta }] = cosec^{2}\theta$

$= R.H.S$

$\displaystyle \therefore L.H.S = R.H.S$

Hence proved

1120284_1140163_ans_29933125f89f44ec89fd1c7f59747e12.jpg

Solve $\sqrt 2 \left( {\sin x + \cos x} \right) = \sqrt 3$

Consider the given equation.

$\sqrt{2}\left( \sin x+\cos x \right)=\sqrt{3}$

$\left( \sin x+\cos x \right)=\dfrac{\sqrt{3}}{\sqrt{2}}$

On squaring both sides, we get

${{\left( \sin x+\cos x \right)}^{2}}=\dfrac{3}{2}$

${{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x=\dfrac{3}{2}$

We know that

${{\sin }^{2}}x+{{\cos }^{2}}x=1$

$\sin 2x=2\sin x\cos x$

Therefore,

$1+\sin 2x=\dfrac{3}{2}$

$\sin 2x=\dfrac{1}{2}$

$\sin 2x=\sin \dfrac{\pi }{6}$

$2x=n\pi +(-1)^n\dfrac{\pi }{6}$

$x=\dfrac{n\pi}{2} +(-1)^n\dfrac{\pi }{12}$

Hence, the value of $x$ is $\dfrac{n\pi}{2} +(-1)^n\dfrac{\pi }{12}$

Find ' $x$ ' if $3\tan^{-1} x + \cot^{-1}x = \pi$ .

$3 \tan^{-1}x+\cot^{-1}x=\pi$

$3 \tan^{-1}x+\dfrac{\pi }{2}-\tan^{-1}x=\pi$

$(\because \tan^{-1}x+\cot^{-1}x=\dfrac{\pi }{2})$

$2 \tan^{-1}x=\pi -\dfrac{\pi }{2}$

$2 \tan^{-1}x=\dfrac{\pi }{2}$

$\tan^{-1}x=\dfrac{\pi }{4}$

$x= \tan \dfrac{\pi }{4}$

$x=1$

Solve ${\tan ^2}\theta + {\cot ^2}\theta = 2$

Consider the given equation.

${{\tan }^{2}}\theta +{{\cot }^{2}}\theta =2$

${{\tan }^{2}}\theta +\dfrac{1}{{{\tan }^{2}}\theta }=2$

${{\tan }^{4}}\theta +1=2{{\tan }^{2}}\theta$

${{\tan }^{4}}\theta -2{{\tan }^{2}}\theta +1=0$

${{\left( {{\tan }^{2}}\theta -1 \right)}^{2}}=0$

${{\tan }^{2}}\theta -1=0$

${{\tan }^{2}}\theta =1$

${{\tan }^{2}}\theta ={{\tan }^{2}}\dfrac{\pi }{4}$

$\theta =n\pi \pm \dfrac{\pi }{4}$

Hence, this is the answer.

Solve $\tan 5\theta = \cot 2\theta$

We have,

$\tan 5\theta =\cot 2\theta$

$\dfrac{\sin 5\theta }{\cos 5\theta }=\dfrac{\cos 2\theta }{\sin 2\theta }$

$\sin 5\theta \sin 2\theta =\cos 5\theta \cos 2\theta$

$\cos 5\theta \cos 2\theta -\sin 5\theta \sin 2\theta =0$

$\cos \left( 5\theta +2\theta \right)=0$

$\cos 7\theta =\cos \dfrac{\pi }{2}$

$7\theta =2n\pi \pm \dfrac{\pi }{2}$

$\theta =\dfrac{2n\pi }{7}\pm \dfrac{\pi }{14}$

Hence, this is the answer.

If $\tan \theta=\dfrac{3}{A}$ , find $\cos \theta \csc \theta$ .

We have,

$\tan \theta =\dfrac{3}{A}$

Since,

$\cos \theta \csc \theta$

$\Rightarrow \cos \theta \times \dfrac{1}{\sin \theta }$

$\Rightarrow \dfrac{\cos \theta }{\sin \theta }$

$\Rightarrow \cot \theta$

$\Rightarrow \dfrac{1}{\tan \theta }$

$\Rightarrow \dfrac{1}{\dfrac{3}{A}}$

$\Rightarrow \dfrac{A}{3}$

Hence, this is the answer.

Solve the equations
$3(\sec^{2} \theta + \tan^{2} \theta) = 5$ .

Consider the given equation.

$3\left( {{\sec }^{2}}\theta +{{\tan }^{2}}\theta \right)=5$

${{\sec }^{2}}\theta +{{\tan }^{2}}\theta =\dfrac{5}{3}$

We know that

${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta$

Therefore,

$1+{{\tan }^{2}}\theta +{{\tan }^{2}}\theta =\dfrac{5}{3}$

$2{{\tan }^{2}}\theta =\dfrac{5}{3}-1$

$2{{\tan }^{2}}\theta =\dfrac{2}{3}$

${{\tan }^{2}}\theta =\dfrac{1}{3}$

$\tan \theta =\dfrac{1}{\sqrt{3}}$

$\tan \theta =\tan \dfrac{\pi }{6}$

$\theta =n\pi +\dfrac{\pi }{6}$

Hence, this is the answer.

If $co\sec{A}=2$ , find the value of $\cfrac{1}{\tan{A}}+\cfrac{\sin{A}}{1+\cos{A}}$

$cosec A =2$

$sin A=\dfrac{1}{2}$

$A=30^{o}$

Therefore,

$\dfrac{1}{tan A}+\dfrac{sin A}{1+cos A}$ =

$\dfrac{1}{tan 30^{o}}+\dfrac{sin 30^{o}}{1+cos 30^{o}}$

$=\sqrt{3}+\dfrac{\dfrac{1}{2}}{1+\dfrac{\sqrt{3}}{2}}$

$=\sqrt{3}+\dfrac{1}{2+\sqrt{3}}$

$=\dfrac{2\sqrt{3}+4}{2+\sqrt{3}}$

On Rationalisation,

$=4\sqrt{3}+8-6-4\sqrt{3}$

$=2$

If $\sin{\theta}=\cfrac{1}{2}$ , show that $(3\cos{\theta}-4{\cos}^{3}{\theta})=0$

$\sin\theta =1/2\Rightarrow \theta =\pi /6$

$3\cos\theta -4\cos^{3}\theta =\cos3\theta$

$=\cos3\dfrac{\pi }{6}$

$=\cos\pi /2=0$

Write the complementary angle of $\dfrac{1}{3}$ of $180^o$ .

$\dfrac13*180^o=60^o$

Its Complimentary angle will be

$=90^o-60^o=30^o$

Prove that $\cos 4x=1-8\sin^2 x \cos^2x$

Identities:

$\cos 2\theta = 1-2\sin^2\theta$ (1)

$\sin 2\theta = 2\sin\theta\cos\theta$ (2)

using (1) :-

$\cos 4x$ =

$1-2(\sin 2x)^2$

using (2) :-

$=1-2(2\sin x\cos x)^2$

$=1-8\sin^2 x\cos^2 x$

Hence, proved.

cos $x=\frac{1}{2}$
Find the principle solutions.

$\therefore cos x = \left ( \dfrac{\pi }{3} \right )$ and cos x =

$cos\left ( 2\pi- \dfrac{\pi }{3} \right )$

$= cos \left ( \dfrac{5\pi }{3} \right )$ such that

$0< \dfrac{\pi }{3}< 2\pi$ and

$0< \dfrac{5\pi }{3}< 2\pi$

$\therefore$ The required principal solutions are

$x=\dfrac{\pi }{3}$ and

$x=\dfrac{5\pi }{3}$

If $\sin A = \frac{1}{3}$ , then find the value of ${\mathop{\rm cosAcosec}\nolimits} + tanAsecA$

$sin A= \dfrac{1}{3}\Rightarrow cos A= \sqrt{1-sin^{2}A}= \sqrt{1-\dfrac{1}{9}}=\sqrt{\dfrac{8}{9}}=\dfrac{2\sqrt{2}}{3}$

$tan A= \dfrac{sin A}{cos A}= \dfrac{\dfrac{1}{3}}{\dfrac{2\sqrt{2}}{3}}= \dfrac{1}{2\sqrt{2}}$

$sec A= \dfrac{1}{cos A}= \dfrac{3}{2\sqrt{2}}$

$cosec A= \dfrac{1}{sin A}= 3$

$cos A. cosec A+ tan A. sec A$

$=\dfrac{2\sqrt{2}}{3}.3+\dfrac{1}{2\sqrt{2}}.\dfrac{3}{2\sqrt{2}}$

$=2\sqrt{2}+\dfrac{3}{8}=\dfrac{3+16\sqrt{2}}{8}$

1115876_1142589_ans_a18b980424744deb94765b0ad75fd76d.jpg

Find number of solutions of equation $\sin x=-4x+1$

It is clearly evident from above graph that there is only one solution for given equation .

1225460_1154967_ans_c494297af12044bbaac63032edfb68e5.jpg

1225460_1154967_ans_c494297af12044bbaac63032edfb68e5.jpg

Find the principal solution of the following equation:
$\cot { x=\sqrt { 3 } }$

1512980_1147259_ans_be192bd49bf6493095179f414a3cbbe4.jpg

Value of $\sin ^ { 2 } A + \cos ^ { 2 } A = ?$

$Sin^2A+Cos^2A = (\dfrac{p}{h})^2+(\dfrac{b}{h})^2 = \dfrac{p^2+b^2}{h^2} = 1$

from Pythagoras theorem

$h^2=p^2+b^2\\\therefore Sin^2A+Cos^2A = 1$

Find the value of ${ \left( \cfrac { \cos { A } +\cos { B } }{ \sin { A } -\sin { B } } \right) }^{ n }+{ \left( \cfrac { \sin { A } +\sin { B } }{ \cos { A } -\cos { B } } \right) }^{ n }$ (where $n$ is an even)

$\displaystyle (\frac{cosA+cosB}{sinA-sinB})^{n}+(\frac{sinA+sinB}{cosA-cosB})^{n}$

$\displaystyle = (\frac{2cos(\frac{A+B}{2})cos(\frac{A-B}{2})}{2cos(\frac{A-B}{2})sin(\frac{A-B}{2})})^{n}+(\frac{2sin(\frac{A+B}{2})cos(\frac{A-B}{2})}{-2sin(\frac{A+B}{2})sin(\frac{A-B}{2})})^{n}$

$\displaystyle = cot^{n}(\frac{A-B}{2})+(-1)^{n}cot^{n}(\frac{A-B}{2})$

if n is even, than

$(-1)^{n} = 1$ then the given expression becomes

$\displaystyle 2cot^{n}(\frac{A-B}{2})$

If n is odd then

$(-1)^{n} = -1$ then the given expression becomes 0

1077098_1183593_ans_e46346d0a2b044f1b7072aac0d1e4721.png

If $0 < \theta < 90^o$ and $\sec \theta = \text{cosec}60^o$ , find the value of $2{\cos ^2}\theta - 1$

$0< \theta < 90^{\circ}$

$sec\theta = cosec60$

$\Rightarrow sec\theta = \frac{1}{sin60} = \frac{2}{\sqrt{3}}$

$\Rightarrow \frac{1}{cos\theta } = \frac{2}{\sqrt{3}} \Rightarrow cos\theta = \frac{\sqrt{3}}{2} \Rightarrow \theta = 30^{\circ}$

$2cos^{2}\theta -1 = 2(cos30)^{2}-1$

$\Rightarrow 2\times (\frac{\sqrt{3}}{2})^{2}-1$

$\Rightarrow 2\times \frac{3}{4}-1\Rightarrow \frac{3}{2}-1= \frac{1}{2}$

$\therefore 2cos^{2}\theta -1 = \frac{1}{2}$

1112250_1191150_ans_a36939fe33254ebfb7b5309e6d8b319f.jpeg

$tan^{-1}\dfrac{1}{3}+tan^{-1}\dfrac{1}{7}+tan^{-1}\dfrac{1}{5}+tan^{-1}\dfrac{1}{8}=\dfrac{\pi }{4}.$

$tan^{-1}(\frac{1}{3})+tan^{-1}(\frac{1}{7})+tan^{-1}(\frac{1}{5})+tan^{-1}(\frac{1}{8})$

$\Rightarrow tan^{-1}(\frac{10/21}{20/21})+tan^{-1}(\frac{13/40}{39/40})$

$\Rightarrow tan ^{-1}(\frac{1}{2})+tan^{-1}(\frac{1}{3})$

$\Rightarrow tan^{-1}(\frac{5/6}{5/6})$

$\Rightarrow tan^{-1}(1)$

$\Rightarrow \frac{\pi }{4}$

$tan^{-1}(A)+tan^{-1}B =$

$tan^{-1}(\frac{A+B}{1-AB})$

1113339_1204173_ans_6b43d38fdd15465899c6bf1bd02dc027.jpg

Prove that: $\displaystyle \frac { \sec \theta + \tan \theta - 1 } { \tan \theta - \sec \theta + 1 } = \frac { \cos \theta } { 1 - \sin \theta }$

Put

$1=\text{sec}^2\theta-\tan^2\theta=(\text{sec}\theta-\tan\theta)(\text{sec}\theta+\tan\theta)$

$\implies \dfrac{\text{sec}\theta+\tan\theta-(\text{sec}\theta-\tan\theta)(\text{sec}\theta+\tan\theta)}{\tan\theta-\text{sec}\theta+1}$

$\implies \dfrac{\text{sec}\theta+\tan\theta (1-\text{sec}\theta+\tan\theta)}{(1-\text{sec}\theta+\tan\theta)}$

$\implies \text{sec}\theta+\tan\theta$

Put

$\text{sec}\theta = \dfrac{1}{\cos\theta}$ and

$\tan\theta =\dfrac{\sin\theta}{\cos\theta}$

$\implies \dfrac{1}{\cos\theta}+\dfrac{\sin\theta}{\cos\theta}$

Multiply and divide by

$1-\sin\theta$

$\implies \dfrac{1-\sin^2\theta}{\cos\theta(1-\sin\theta)}$

$\implies \dfrac{\cos\theta}{1-\sin\theta}$

Solve: $\dfrac{1+tan^2\theta}{1-tan^2\theta}=sec2\theta.$

$\dfrac{1+tan^2\theta}{1-tan^2\theta}=\dfrac{1+\dfrac{\sin^2\theta}{\cos^2\theta}}{1-\dfrac{\sin^2\theta}{\cos^2\theta}}=\dfrac{\cos^2\theta+\sin^2\theta}{\cos^2\theta-\sin^2\theta}=\dfrac{1}{\cos2\theta}=\sec2\theta$

Prove : $\dfrac{\cot A +\text{ cosec} A -1}{\cot A - \text{cosec} A +1} = \dfrac{1 +\cos A}{\sin A}$

$\cfrac{\cot A+cosec A-1}{\cot A-cosec A+1}=\cfrac{1+\cos A}{\sin A}\\=\cfrac{\cot A+cosec A-(cosec^2A-\cot^2 A)}{\cot A-cosec A+1}[\because cosec^2 A-\cot^2A=1]\\=\cfrac{(\cot A+cosec A)-(\cot A+cosec A)(cosec A-\cot A)}{\cot A-cosec A+1}\\=\cfrac{(\cot A+cosec A)(1-cosec A+\cot A)}{\cot A-cosec A+1}\\ =\cfrac{\cos A}{\sin A}=\cfrac{1}{\sin A}\\=\cfrac{\cos A+1}{\sin A}\\=\cfrac{1+\cos A}{\sin A}=RHS$

Prove $\dfrac{1}{\sec\theta -1}-\dfrac{1}{\sec\theta +1}=2\cot^2\theta$

$LHS=\dfrac{1}{\text{sec}\theta-1}-\dfrac{1}{\text{sec}\theta+1}=\dfrac{\text{sec}\theta+1-\text{sec}\theta+1}{\text{sec}^{2}\theta-1}=\dfrac{2}{\tan^2 \theta}=2\cot^2 \theta=RHS$

$\because L.H.S=R.H.S$

$\dfrac{1}{\sec\theta -1}-\dfrac{1}{\sec\theta +1}=2\cot^2\theta$

Hence Proved

Solve $\dfrac{\cos{\theta}}{1-\sin{\theta}}+\dfrac{\cos{\theta}}{1+\sin{\theta}}=4$ for $\theta <90$

solve

$\frac{cos\theta }{1-sin\theta }+\frac{cos\theta }{1+sin\theta } = 4$

$\theta < 90$

$\frac{cos\theta }{1-sin\theta }+\frac{cos\theta }{1+sin\theta } = 4$

$\frac{cos\theta (1+sin\theta )+cos\theta (1-sin\theta )}{1-sin^{2}\theta } = 4$

$\frac{cos\theta +cos\theta .sin\theta +cos\theta -cos\theta sin\theta }{cos^{2}\theta } = 4$

$\therefore \frac{2cos\theta }{cos^{2}\theta } = 4$

$\therefore \frac{2}{4} = cos\theta$

$\therefore cos\theta = \frac{1}{2}$

$\therefore \theta = 60^{\circ}$

1122042_1248008_ans_95c2f8f3b2f349ca8d2bf2220d52e43f.jpg

If $\cos \theta=\dfrac{3}{5}$ , find the value of $\cot \theta +cosec \ \theta$ .

1339270_1264355_ans_1fbb51b7fecc42b695160cd54de1739e.jpg

If $\sin \theta =\dfrac{a}{\sqrt{a^2+b^2}}$ , $0<\theta <90^o$ , find the values of $\cos \theta$ and $\tan \theta$ .

1339267_1261618_ans_3bdd2664038f4f81aa212c02645ee135.jpg

The value of $\cot^{2}C-\dfrac{1}{\sin^{2}C}$ is

$\cot ^{ 2 }{ C } -\cfrac { 1 }{ \sin ^{ 2 }{ C } } \\ =\cfrac { \cos ^{ 2 }{ C } }{ \sin ^{ 2 }{ C } } -\cfrac { 1 }{ \sin ^{ 2 }{ C } } \\ =\cfrac { -(1-\cos ^{ 2 }{ C } ) }{ \sin ^{ 2 }{ C } } \\ =\cfrac { -\sin ^{ 2 }{ C } }{ \sin ^{ 2 }{ C } } \\ =-1$

The value of $\tan^{2}A-\sec^{2}A$ is

$\tan ^{ 2 }{ A } -\sec ^{ 2 }{ A } =-(\sec ^{ 2 }{ A } -\tan ^{ 2 }{ A } )=-1$

Write $\tan\theta$ in terms of $\sin\theta$ .

We have

$\tan \theta=\cfrac{\sin \theta}{\cos \theta}$

As we know

$\cos \theta=\sqrt{1-\sin^2\theta}$

$\implies \tan\theta=\cfrac{\sin\theta}{\sqrt{1-\sin^2\theta}}$

If $\sin 3A=\cos (A-10^o)$ , then find the value of A, where 3A is an acute angle.

1338270_1252703_ans_01c86476def348689b7639f74cc5010f.jpg

Prove that $\dfrac {\cot A-\cos A}{\cot A+\cos A}=\dfrac {\text{cosec} A-1}{\text{cosec} A+1}$

1209627_1249209_ans_abc04a1a0a0f4f8087b7ca639de5d5c0.jpg

Eliminate $\theta$ , if $x=a\cos\theta, y=a\sin \theta$ .

$x=a\cos\theta;y=a\sin\theta\\x^2+y^2=a^2(\cos^2\theta+\sin^2\theta)\\x^2+y^2=a^2$

Prove that :
$(\csc A-\sin A)(\sec A-\cos A)=\dfrac {1}{\tan A+\cot A}$

$L.H.S = (cosec A- sin A) (sec A- cos A)$

$=(\frac{1}{sin A}-sin A)(\frac{1}{cos A}-cos A)=\frac{(1-sin^{2}A)(1- cos^{2}A)}{sin A cos A}$

$=\frac{cos^{2}A\times sin^{2}A}{sin A cos A}= sin A cos A$

R.H.D=

$\frac{1}{tan A+ cot A}=\frac{1}{\frac{sin A}{cos A}+\frac{cos A}{sin A}}=\frac{sin A cos A}{sin^{2}A+cos^{2}A} = sin A cos A$

$\Rightarrow L.H.S = R.H.S$

1194256_1379950_ans_7da9f2b315fe46d3ade4dc6a0b6f1b03.jpg

If $7\csc\alpha-3\cot\alpha=7$ prove that $7\cot\alpha-3\csc\alpha=3$

A] Given

$7cosec \theta - 3 cot \theta = 7$

squaring,

$\Rightarrow (7 cosec \theta -3 cot \theta )^{2} = 49$

$\Rightarrow 49 cosec^{2} \theta + 9 cot^{2} \theta - 2 \times 7 \times 3 cosec \theta cot \theta = 49$

$\Rightarrow 49 (1 + cot 2 \theta) + 9 (cosec2\theta - 1) - 42 cosec \theta cot\theta=49$

$\Rightarrow 49+49cot2\theta+9cosec2\theta-9 - 42 cosec \theta cot\theta=49$

$\Rightarrow \Rightarrow (7cot\theta)^{2} + (3cosec \theta)^{2} -2 \times 7 \times 3 cot \theta cosec \theta = 9$

$\Rightarrow (7 cot\theta -3cosec\theta )^{2} = 9$

$\Rightarrow 7 cot \theta - 3 cosec \theta = 3$

Hence proved

1193903_1295603_ans_987f5bf5b16947f08057b9e8c10d795e.jpg

Prove that $sec^{4}\theta-tan^{4}\theta=1+2 tan^{2}\theta.$

LHS

$\sec^4\theta -\tan^4\theta$

$\Rightarrow (\sec^2\theta -\tan^2\theta)(\sec^2\theta +\tan^2\theta)$

$[\because a^2-b^2=(a-b)(a+b)]$

$\Rightarrow 1(\sec^2\theta +\tan^2\theta)$

$[\because1+\tan^2\theta =\sec^2\theta\Rightarrow \sec^2\theta -\tan^2\theta =1]$

$\Rightarrow (1+\tan^2\theta +\tan^2\theta)$

$\Rightarrow 1+2\tan^2\theta =$ RHS.

Hence proved.

Find the principal solution or solutions of $\sin x = \dfrac { 1 } { \sqrt { 2 } }$

$\sin x=\dfrac{1}{\sqrt{2}}$

$x=\dfrac{\pi}{4}$ and

$\dfrac{3\pi}{4}$ are principal solutions.

Find the value of $\sin (270^o)$ .

Now,

$\sin (270^o)$

$=\sin(3\times 90^o+0^o)$

$\sin (270^o)$

$=-\cos 0^o$

$\sin (270^o)$

$=-1$ .

Prove $\dfrac{1-\cos A}{1+\cos A}=(\csc A-\cot A)^2$

$\dfrac{1-\cos A}{1+\cos A}\times \dfrac{1-\cos A}{1-\cos A}=\dfrac{1+\cos^2A-2\cos A}{1-\cos^2A}=\dfrac{1+\cos^2A-2\cos A}{\sin^2A}$

$=cosec^2A+\cot^2A-2\cot A cosec A$

$=(cosec A-\cot A)^2$ .

1216087_1396874_ans_0c213925b7b741b09a2cfd1f915ab80c.jpg

If $\sec\theta+\tan\theta=p$ , find the value of $\csc\theta$

$\sec 2\theta - \tan 2\theta = 1$

$(\sec \theta - \tan \theta )P = 1$

$\sec \theta - \tan \theta = \dfrac{1}{p}$ ...(1)

$\sec \theta + \tan \theta = p$ ...(2)

eqn. (1) + (2)

$\sec \theta = \dfrac{1}{2} \left(p + \dfrac{1}{p}\right)$ ...(3)

eqn. (1) - (2)

$\tan \theta = -\dfrac{1}{2} \left(\dfrac{1}{p}-p\right)$ ...(4)

eqn. (3)

$\div$ (4)

$cosec \theta = \dfrac{p^2 + 1}{p^2 - 1}$

1226980_1398580_ans_efd2a9b1a24a4462a4da331c0e9031d9.jpg

Make the following expression a perfect square:
$1+2\sin \theta.\cos \theta$ .

Now,

$1+2\sin \theta.\cos \theta$

$=\sin^2 \theta+\cos^2 \theta+2\sin \theta.\cos \theta$ [ Since

$\sin^2 \theta+\cos^2\theta=1$ ]

$=(\sin \theta+\cos \theta)^2$ .

This is the required perfect square term.

Evaluate: $3\cot^{2}\theta-3\text{cosec}^{2}\theta$

$3\cot^2\theta -3cosec^2\theta =-3(cosec^2\theta -\cot^2\theta)$

$=-3$

$\{cosec^2\theta -\cot^2\theta =1\}$ .

1208059_1395337_ans_ca0ca2c0af994ae08296d2262da25951.jpg

Prove that
$\frac{cosA}{1-tanA}+\frac{sinA}{1-cot A}=sinA +cosA$

$LHS=$

$\dfrac{CosA}{1-TanA}+\dfrac{SinA}{1-CotA}=\dfrac{CosA}{1-\dfrac{SinA}{CosA}}+\dfrac{SinA}{1-\dfrac{CosA}{SinA}}=\dfrac{Cos^2{A}-Sin^2{A}}{CosA-SinA}=CosA+SinA$

$=RHS$

Prove
$tan \Theta +sec\theta=\frac{1}{sec\Theta -tan\Theta }$

$\tan\theta +\sec\theta =\dfrac{1}{\sec\theta -\tan\theta}$

Solving LHS

$=\tan\theta +\sec\theta$ (on rationalising)

$\Rightarrow \dfrac{(\tan\theta +\sec\theta)(\tan\theta -\sec\theta)}{\tan\theta -\sec\theta}=\dfrac{\tan^2\theta -\sec^2\theta}{\tan\theta -\sec\theta}$

We know

$1+\tan^2\theta =\sec^2\theta$

$\Rightarrow \tan^2\theta -\sec^2\theta =-1$

Then

$\Rightarrow \dfrac{-1}{\tan\theta -\sec\theta}$

$\Rightarrow \dfrac{1}{\sec\theta -\tan\theta}=$ RHS.

1204171_1393329_ans_4a88f3c83e1948d78f30173b41b07811.JPG

The value of x fo which $\sin(\pi x)+\cos (\pi x)=0$

$\sin \pi x+\cos \pi x=0\\\sin \pi x=-\cos \pi x\\-\tan \pi x=1\\\tan \pi x=-1\\\tan \pi x=\tan \dfrac{3\pi}{2}\\x=\dfrac{3\pi}{2}$

Prove that : $(1+\sin\theta-\cos\theta)^{2}+(1-\sin\theta-\cos\theta)^{2}=4(1-\cos\theta)$

$[(1-\cos\theta)+\sin\theta]^2=[(1-\cos\theta)-\sin\theta)^2$

$(a+b)^2+(a-b)^2=2(a^2+b^2)=2(1+\cos^2\theta -2\cos\theta +\sin^2\theta)$

$=2(1+1-2\cos\theta)=4(1-\cos\theta)$ .

1214906_1396577_ans_4104ac72a27247f595a7160f2531d60b.jpg

If $\tan 9\theta=\cot \theta$ , where $9\theta < 90^{o}$ then find the value of $\theta$ .

Given

$\tan 9\theta=\cot \theta$ and

$9\theta<90^{\circ}\implies \theta<10^{\circ}$

$\tan 9\theta=\tan (90^{\circ}-\theta)$

$\implies 9\theta=90^{\circ}-\theta$

$\implies 10\theta=90^{\circ}$

$\implies \theta=9^{\circ}$

Find the degree measure of ${\dfrac{1}{4}}^{c}$

We have

${\pi}^{c}={180}^{\circ}$

$\therefore\,{1}^{c}={\dfrac{180}{\pi}}^{\circ}$

${\dfrac{1}{4}}^{c}={\left(\dfrac{1}{4}\times \dfrac{180}{\pi}\right)}^{\circ}$

$={\left(\dfrac{1}{4}\times 180\times\dfrac{7}{22}\right)}^{\circ}$

$={\left(\dfrac{315}{22}\right)}^{\circ}$

$={14}^{\circ}{\left(\dfrac{7}{22}\times 60\right)}^{\prime}$

$={14}^{\circ}{\left(\dfrac{7}{11}\times 30\right)}^{\prime}$

$={14}^{\circ}{\left(19\dfrac{1}{11}\right)}^{\prime}$

$={14}^{\circ}{19}^{\prime}{\left(\dfrac{1}{11}\right)}^{\prime}$

$={14}^{\circ}{19}^{\prime}{\left(\dfrac{1}{11}\times 60\right)}^{\prime\prime}$

$={14}^{\circ}{19}^{\prime}{5}^{\prime\prime}$

Prove that : $\dfrac { 1 + \sin \theta } { \cos \theta } + \dfrac { \cos \theta } { 1 + \sin \theta } = 2 \sec \theta$

$\frac{1+ sin\theta }{cos \theta }+\frac{cos\theta }{1+ sin\theta }=2 sec\theta$

$\frac{(1+sin\theta )^{2}+cos^{2}\theta }{cos \theta (1+ sin \theta )}=2 sec \theta$

$\frac{1+ sin^{2} \theta +2 sin \theta +cos^{2} \theta }{cos \theta (1+ sin \theta )}=2 sec \theta$

$sin^{2} \theta +cos^{2} \theta =1$

$\frac{1+1+2 sin \theta }{cos \theta (1+ sin \theta )}= 2 sec \theta$

$\frac{2+2 sec \theta }{cos \theta (1+ sin \theta )}= 2 sec \theta$

$\frac{2(1-1 sin \theta )}{cos \theta (1+ sin \theta )}= 2 sec \theta$

$2 sec \theta = 2 sec \theta$

$L.H.S=R.H.S$

1226200_1502817_ans_b0778e4c69eb495388810e41b2f7dabe.jpg

If $\tan (A+B) =\sqrt{3}$ and $\tan(A-B)=\dfrac{1}{\sqrt{3}}$ find $A$ and $B$ . A and B are acute angles. Then A+B in degrees

Given

$\tan (A+B) =\sqrt{3}\implies A+B=60^{\circ}\cdots(1)$

$\tan(A-B)=\dfrac{1}{\sqrt{3}}\implies A-B=30^{\circ}\cdots(2)$

$(1)+(2)\implies 2 A=90^{\circ}\implies A=45^{\circ}$

Put

$A$ value in

$(1)$

$\implies B=60^{\circ}-45^{\circ}=15^{\circ}$

$sec\Theta +tan\Theta =p$ find $cosec\Theta$

Given

$\text{sec}\theta+\tan \theta=p\dots(1)$

As we know that

$\text{sec}^2 \theta-\tan^2 \theta=1$

$\implies (\text{sec}\theta+\tan \theta)(\text{sec}\theta-\tan \theta)=1$

$\implies \text{sec}\theta-\tan \theta=\dfrac{1}{p}\cdots(2)$

$(1)+(2)\implies 2\text{sec}\theta=p+\dfrac{1}{p}\implies \text{sec}\theta=\dfrac{1}{2}\bigg(p+\dfrac{1}{p}\bigg)$

$(1)-(2)\implies 2\tan \theta=p-\dfrac{1}{p}\implies \tan \theta=\dfrac{1}{2}\bigg(p-\dfrac{1}{p}\bigg)$

$\text{cosec}\theta=\dfrac{\text{sec}\theta}{\tan \theta}=\dfrac{\frac{1}{2}(p+\frac{1}{p})}{\frac{1}{2}(p-\frac{1}{p})}=\dfrac{\frac{p^2+1}{p}}{\frac{p^2-1}{p}}=\dfrac{p^2+1}{p^2-1}$

$\frac { \cos { A } -\sin { A+1 } }{ \cos { A } +\sin { A-1 } } =cosecA+\cot { A }$

$LHS=\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{(\cos A-\sin A)+1}{(\cos A+\sin A)-1}\times \dfrac{(\cos A+\sin A)+1}{(\cos A+\sin A)+1}$

$=\dfrac{(\cos A+\sin A)(\cos A-\sin A)+(\cos A+\sin A)+(\cos A-\sin A)+1}{(\cos A+\sin A)^2-1}$

$=\dfrac{\cos^2 A-\sin^2 A+2\cos A+1}{\cos^2 A+\sin^2 A+2\sin A\cos A-1}$

$=\dfrac{2\cos^2 A+2\cos A}{2\sin A\cos A}$

$=\cot A+\text{cosec}A=RHS$

Find the degree measure of ${\dfrac{\pi}{8}}^{c}$

We have

${\pi}^{c}={180}^{\circ}$

$\therefore\,{1}^{c}={\dfrac{180}{\pi}}^{\circ}$

${\dfrac{\pi}{8}}^{c}={\left(\dfrac{\pi}{8}\times \dfrac{180}{\pi}\right)}^{\circ}={\left(22\dfrac{1}{2}\right)}^{\circ}={22}^{\circ}\times \dfrac{1}{2}\times {60}^{\circ}={22}^{\circ}{30}^{\prime}$

Express the following degrees into grades (i.e., sexagesimal system to centesimal system)
$63^{\circ}$

We know that to change degree into grades, add one-ninth into degrees.

$\therefore$

$63^{\circ} = 63 + \dfrac{1}{9} \times 63$

$= 63 + 7 = 70^{g}$

If $\tan { \theta } +\sec { \theta } =x$ , show that $\sin { \theta } =\cfrac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 }$

$\tan { \theta } +\sec { \theta } =x\Rightarrow \tan { \theta } =x-\sec { \theta }$
squaring both sides

$\Rightarrow \tan ^{ 2 }{ \theta } ={ (x-\sec { \theta } ) }^{ 2 } \\ \Rightarrow \tan ^{ 2 }{ \theta } ={ x }^{ 2 }+\sec ^{ 2 }{ \theta } -2x\sec { \theta } \\ \Rightarrow \sec ^{ 2 }{ \theta } -1={ x }^{ 2 }+\sec ^{ 2 }{ \theta } -2x\sec { \theta } \left[ \because 1+\tan ^{ 2 }{ \theta } =\sec ^{ 2 }{ \theta } \right]$

$\Rightarrow -1-{ x }^{ 2 }--2x\sec { \theta } \Rightarrow \sec { \theta } =\cfrac { 1+{ x }^{ 2 } }{ 2x }$

$\tan { \theta } =x-\sec { \theta } \Rightarrow \cfrac { \sin { \theta } }{ \cos { \theta } } =x-\sec { \theta } \\ \Rightarrow \sin { \theta } \left( \cfrac { 1+{ x }^{ 2 } }{ 2x } \right) =x-\left( \cfrac { 1+{ x }^{ 2 } }{ 2x } \right) \\ \Rightarrow \sin { \theta } \left( \cfrac { 1+{ x }^{ 2 } }{ 2x } \right) =\cfrac { 2{ x }^{ 2 }-1+{ x }^{ 2 } }{ 2x } \\ \Rightarrow \sin { \theta } =\cfrac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 } =RHS$
hence proved

Convert $\dfrac{\pi}{12}$ radians into degree (sexagesimal system).

$\because$

$1$ radian

$= \dfrac{180}{\pi}$ degrees

$\dfrac{\pi}{12}$ radians

$= \dfrac{180}{\pi} \times \dfrac{\pi}{72}$ degree

$= \left ( \dfrac{5}{2} \right )^{\circ} = 2^{\circ} 30'$

Find the principal solution of the following equation:
$cos \theta=\frac{1}{2}$

1888248_1846121_ans_70c922c3ed5f4848a63bdb649189541e.jpeg

If $A - B = C$ and $A + B =\frac{\pi }{2}$ then $\tan A = \tan B + 2\tan C$ .

$\because A + B =\dfrac{\pi }{2} \Rightarrow \tan A\tan B = 1$
and

$\tan C = \tan(A - B) =\dfrac{\tan A - \tan B}{1+ \tan A\tan B}$

$\Rightarrow 2\tan C = \tan A - \tan B$

State true(1) or false(0):
$\left ( \tan \theta +2 \right )\left ( 2 \tan \theta +1 \right )=5 \tan\theta +sec^{2} \theta$

LHS

$=2\tan^2\theta+\tan\theta+4\tan\theta+2=2\tan^2\theta+5\tan\theta+2$

$=2\sec^2\theta-2+5\tan\theta+2$

$=2\sec^2\theta+5\tan\theta$

$\neq$ RHS

Fill in the blank
$\sec A(1-\sin A)(\sec A + \tan A)=$ _____

Given : $\sec A(1-\sin A)(\sec A + \tan A)$

$\dfrac{1}{\cos A} {(1-\sin A)} \left (\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\right )$ ................... $\left (\because sec A=\dfrac1{cos A} \right )$

$\left (\dfrac{1-\sin A}{\cos A}\right ) \left (\dfrac{1+\sin A}{\cos A} \right )$

$\dfrac{1-\sin^2A}{\cos^2A}$

As, $1-\sin^2A =\cos^2A$

$\dfrac{ \cos^2A }{\cos^2A}$ $=1$

The number of solutions of $3 sec\theta-5=4 tan \theta$ in $[0, 4\pi]$ be $k$ .Find $k$ ?

Convert into

$sin\theta$ &

$cos\theta$ , we get

$4 sin\theta+5 cos \theta=3, cos\theta \neq 0$ ,
Solved

$\displaystyle \dfrac {4}{\sqrt {41}}=cos\beta, sin\alpha = \dfrac{5}{41}$

$\theta-\alpha=2n\pi \pm \beta$

$\therefore \theta=2n\pi + \alpha +\beta$ now in

$[0, 2\pi], \theta=2n\pi+(\alpha+\beta)$ Given two

$sol^n$
or

$=2\pi +\alpha -\beta$

$\therefore$ Total No. of

$sol^n.=8$

The number of values of $x$ between $0$ and $2\pi$ that satisfies the equation $\sin x+ \sin 2x+ \sin 3x=\cos x+ \cos 2x+ \cos 3x$ must be-

Given that

sin x + sin 2x + sin 3x = cos x + cos 2x + cos 3x

(sin 3x + sin x) + sin 2x = ( cos 3x + cos x ) + cos 2x

2[sin(3x+x)/2].[cos(3x-x)/2] + sin 2x = 2[cos(3x+x)/2].[cos(3x-x)/2] + cos 2x

2[sin(4x)/2].[cos(2x)/2] + sin 2x = 2[cos(4x)/2].[cos(2x)/2] + cos 2x

2 sin 2x . cos x + sin 2x = 2 cos 2x . cos x + cos 2x

2 sin 2x . cos x + sin 2x - 2 cos 2x . cos x - cos 2x = 0

sin 2x ( 2 cos x + 1) - cos 2x (2 cos x + 1) = 0

( 2 cos x + 1) ( sin 2x - cos 2x ) = 0

⇒

2 cos x + 1 = 0 And sin 2x - cos 2x = 0

2 cos x = - 1 And sin 2x = cos 2x

cos x = - 1/2 And sin 2x/cos 2x = cos 2x/cos 2x

cos x = - 1/2 And tan 2x = 1

For cos x = - 1/2 :-

The Reference Angle is π/3

Cos is -ve in Quadrant II & III

For Quad II :- x = π - π/3 = 2π/3

For Quad III :- x = π + π/3 = 4π/3

For tan 2x = 1 :-

The Reference Angle is π/4

tan is +ve in Quadrant I & III

For Quad I :- 2x = π/4 ⇒ x = π/8

For Quad III :- 2x = π + π/4 = 5π/4 ⇒ x = 5π/8

The Answer is 2π/3 , 4π/3 , π/8 and 5π/8 .

$\left ( \cos A-\cos B \right )^{2}+\left ( \sin A-\sin B \right )^{2}=c\sin ^{2}\left ( A-B \right )/2$
Find $2c$

L.H.S.

$=\left ( \cos A-\cos B \right )^{2}+\left ( \sin A-\sin B \right )^{2}$

$=\cos ^{2}A+\cos ^{2}B-2\cos A\cos B+\sin ^{2}A+\sin ^{2}B-2\sin A\sin B$

$=\left ( \cos ^{2}A+\sin ^{2}A \right )+\left ( \cos ^{2}B+\sin ^{2}B \right )-2\left ( \cos A\cos B+\sin A\sin B \right )$

$=2-2\cos \left ( A-B \right )=2\left ( 1-\cos \left ( A-B \right ) \right )=4\sin ^{2}\left ( A-B \right )/2$

Therefore,

$2c=8$

Ans: 8

$\sqrt{\sin ^{4}x+4\cos ^{2}x}-\sqrt{\cos ^{4}x+4\sin ^{2}x}=\cos sx$ Find s

The given expression is equal to

$\sqrt{\sin ^{4}x+4\left ( 1-\sin ^{2}x \right )}-\sqrt{\cos ^{4}x+4\left ( 1-\cos ^{2}x \right )}$

$=\sqrt{\left ( 2-\sin ^{2}x \right )^{2}}-\sqrt{\left ( 2-\cos ^{2}x \right )^{2}}$

$=\left ( 2-\sin ^{2}x \right )-\left ( 2-\cos ^{2}x \right )$

$=\cos ^{2}x-\sin ^{2}x=\cos 2x$

Ans: 2

Solution of $\displaystyle 3\left ( \sec ^{2}\theta+\tan^{2}\theta \right )=5$ is $\displaystyle \ \theta =n\pi \pm \frac { \pi }{ m }$ . Find $m$ .

$3\left( \sec ^{ 2 } \theta +\tan ^{ 2 } \theta \right) =5$

$\displaystyle \Rightarrow \sec ^{ 2 } \theta +\tan ^{ 2 } \theta =\frac { 5 }{ 3 }$

$\displaystyle \Rightarrow \tan ^{ 2 }{ \theta } =\frac { 1 }{ 3 } ={ \left( \frac { 1 }{ \sqrt { 3 } } \right) }^{ 2 }$

$\displaystyle \Rightarrow \theta =n\pi \pm \frac { \pi }{ 6 }$

Find the maximum value of $\sqrt{3}\sin x+\cos x$ .

$\displaystyle \sqrt{3}\sin x+\cos x=2\left ( \frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\sin x \right )=2\sin \left ( x+\frac{\pi }{6} \right )$
which is maximum when

$x+\pi /6=\pi /2$ or

$x=60^{0}$ and has maximum value 2

Ans: 2

In quadrilateral $ABCD$ if $\displaystyle \sin \left ( \dfrac{A+B}{2} \right )\cos \left ( \dfrac{A-B}{2} \right )+\sin \left ( \dfrac{C+D}{2} \right )\cos \left ( \dfrac{C-D}{2} \right )=2$ , then the value $\displaystyle \sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}\sin \dfrac{D}{2}$ is $\dfrac{1}{a}$ Find a

$\displaystyle \sin \left ( \frac{A+B}{2} \right )\cos \left ( \frac{A-B}{2} \right )+\sin \left ( \frac{C+D}{2} \right )\cos \left ( \frac{C-D}{2} \right )=2$

$\Rightarrow$

$\displaystyle \frac{1}{2}\left [ \sin A+\sin B+\sin C+\sin D \right ]=2$

$\Rightarrow$

$\sin A+\sin B+\sin C+\sin D=4$

$\therefore$

$\sin A=1,\sin B=1,\sin C=1,\sin D=1$

$\therefore$

$A+B+C+D=360^{0}$ ......As it is a quadrilateral

$\therefore$

$A=B=C=D=90^{0}$

$\Rightarrow$

$\displaystyle \sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\sin \frac{D}{2}=\frac{1}{4}$

Ans:a = 4

If $\displaystyle 0 < x < \frac {\pi}{4}$ and $\displaystyle cos x + sin x = \frac {5}{4}$ , then the value of $\displaystyle 16(cos x - sin x)^2$ is

$\displaystyle \cos { x } +\sin { x } =\frac { 5 }{ 4 } \Rightarrow { \left( \cos { x } +\sin { x } \right) }^{ 2 }=\frac { 25 }{ 16 }$

$\displaystyle \Rightarrow \cos ^{ 2 }{ x } +\sin ^{ 2 }{ x } +2\sin { x } \cos { x } =\frac { 25 }{ 16 }$
Now

${ \left( \cos { x } -\sin { x } \right) }^{ 2 }={ \left( \cos { x } +\sin { x } \right) }^{ 2 }-4\sin { x } \cos { x }$

$\displaystyle =\frac { 25 }{ 16 } -2\left( \frac { 9 }{ 16 } \right) =\frac { 7 }{ 16 }$

$\therefore 16{ \left( \cos { x } -\sin { x } \right) }^{ 2 }=7$

$\displaystyle \frac{1+\sin 2\theta }{1-\sin 2\theta }=\left ( \frac{1+\tan \theta }{1-\tan \theta } \right )^{b}$
.Find $b$

L.H.S.

$\displaystyle =\frac{1+\sin 2\theta }{1-\sin 2\theta }$

$\displaystyle =\frac{\sin ^{2}\theta +\cos ^{2}\theta +2\sin \theta \cos \theta }{\sin ^{2}\theta +\cos ^{2}\theta -2\sin \theta \cos \theta }$

$\displaystyle =\left ( \frac{\sin \theta +\cos \theta }{\sin \theta -\cos \theta } \right )^{2}=\left ( \frac{1+\tan \theta }{1-\tan \theta } \right )^{2}$ (Dividing numerator and denominator by

$\cos \theta$ )

Ans: 2

Find $a$ if, $a\sin 27^{0}=\left ( 5+\sqrt{5} \right )^{1/2}-\left ( 3-\sqrt{5} \right )^{1/2}$

Consier,

$16\sin ^{2}27^{0}$

$=8\left ( 1-\cos 54^{0} \right )$

$\displaystyle =8\left ( 1-\frac{\sqrt{10}-2\sqrt{5}}{4} \right )$

$=2\left ( 4-\sqrt{10}-2\sqrt{5} \right )$

$=8-2\sqrt{10}-2\sqrt{5}$

$=\left ( 5+\sqrt{5} \right )+\left ( 3-\sqrt{5} \right )-2\sqrt{\left ( 5+\sqrt{5} \right )\left ( 3-\sqrt{5} \right )}$

$=\left \{ \sqrt{5+\sqrt{5}}-\sqrt{3-\sqrt{5}} \right \}^{2}$

$\Rightarrow$

$4\sin 27^{0}=\sqrt{5+\sqrt{5}}-\sqrt{3-\sqrt{5}}$

Ans: 4

If $\sin A\, +\, cosec\: A= 2$ ; find the value of $\sin ^{2}A\, +\, cosec^{2}A$ .

$\sin { A } +cosec\;A=2$

Squaring both sides, we get

$\Rightarrow { \left( \sin { A } +cosec\;A \right) }^{ 2 }=4$

$\Rightarrow \sin ^{ 2 }{ A } +cosec^{ 2 }A+2=4$

$\Rightarrow \sin ^{ 2 }{ A } +cosec^{ 2 }A=2$

Hence, the answer is

$2.$

If $\displaystyle 4\cos^{2}A-3=0$ and $\displaystyle 0^{\circ}\leq A \leq 90^{\circ}$ , then find $\angle A$ in degrees.

$4 \cos^2 A = 3$

$\cos^2 = \dfrac{3}{4}$

$\cos A = \dfrac{\sqrt{3}}{2}$ ....(Neglecting the negative value)

$\cos A = \cos 30$

$A = 30^{\circ}$

If $\displaystyle f\left (x \right )= \sin \left ( \log x \right ),$ then show that $\displaystyle f\left (xy\right )+ f\left ( \frac{x}{y} \right )-2f\left ( x \right )\cos \left ( \log y \right )=0$

Consider,

$\displaystyle f\left( xy \right) +f\left( \frac { x }{ y } \right) -2f\left( x \right) \cos \left( \log y \right)$

$=\sin ({ \log { xy } }) +\sin ({ \log { \cfrac { x }{ y } } }) -2\sin( { \log { x } } )\cos\left( \log y \right)$

$\displaystyle =\sin{ \left( \log { x } +\log { y } \right) } +\sin { \left( \log { x } -\log { y } \right) } -2\sin ({ \log { x } }) \cos\left( \log y \right)$

Applying

$\sin C+\sin D=2\sin \cfrac{(C+D)}{2} \cos \cfrac{(C-D)}{2}$ , we get

$\displaystyle =2\sin { \log { x } } \cos\left( \log y \right) -2\sin { \log { x } } \cos \left( \log y \right)$

$=0$

$\displaystyle \cos^2\ 42^{\circ}-\sin^2 48^{\circ}$

$\displaystyle \cos^2\ 42^{\circ}-\sin^2 48^{\circ}$
=

$\displaystyle \cos^2(90^{\circ} - 48^{\circ}) -\sin^2 48^{\circ}$
=

$\displaystyle \sin^2\ 48^{\circ}-\sin^2 48^{\circ}$
=

$0$

$\displaystyle \sin ^2 62^{\circ}-\cos ^2 28^{\circ}$

$\displaystyle \sin ^2 62^{\circ}-\cos ^2 28^{\circ}$
=

$\displaystyle \sin ^2(90^{\circ} - 28^{\circ}) -\cos ^2 28^{\circ}$
=

$\displaystyle \cos ^228^{\circ}-\cos ^2 28^{\circ}$
=

$0$

If $\displaystyle \sin \theta = \dfrac{45}{53}$ , find the value of $\displaystyle cosec^2 \theta - \cot^2 \theta$

$\sin \theta = \dfrac{45}{53}$

$\sin \theta = \dfrac{P}{H} = \dfrac{45}{53}$

$P = 45, H = 53$

By Pythagoras Theorem,

$H^2 = P^2 + B^2$

$53^2 = 45^2 + B^2$

$2809 = 2025 + B^2$

$B^2 = 784$

$B = 28$

Now,

$cosec^2 \theta + \cot^2 \theta = (\dfrac{H}{P})^2 + (\dfrac{B}{P})^2 = (\dfrac{53}{45})^2 - (\dfrac{28}{45})^2$

$\dfrac{53^2 - 28^2}{45^2}$

$\dfrac{45^2}{45^2}$

$1$

figure $\left|\cot ^{2}C\, -\displaystyle \, \frac{1}{\sin ^{2}C}\right|$

Given,

$\triangle ABC$ , A perpendicular from B on AC, let it cut AC at D, such that

$\angle BDA = \angle BDC = 90^{\circ}$

$AD = 3$ ,

$BD = 4$ and

$BC = 12$
In

$\triangle ABD$ ,

$AB^2 = BD^2 + AD^2$

$AB^2 = 3^2 + 4^2$

$AB = 5$
In

$\triangle BCD$

$BC^2 = CD^2 + BD^2$

$12^2 = CD^2 + 4^2$

$CD^2 = 128$

$CD = 8 \sqrt{2}$
Now,

$\left|\cot^2 C - \dfrac{1}{\sin^2 C}\right| = \left|\cot^2 C - \csc^2 C\right| = \left|\left(\dfrac{B}{P}\right)^2 - \left(\dfrac{H}{P}\right)^2\right|$

$=$

$\left|\left(\dfrac{8\sqrt{2}}{4}\right)^2 - \left(\dfrac{12}{4}\right)^2\right|$

$=$

$|8 - 9|$ =

$1$

$\displaystyle 2\sin^2 54^{\circ}-2\cos ^2\ 36^{\circ}+4\sin^2 30^{\circ}$

$\displaystyle 2\sin^2 54^{\circ}-2\cos ^2\ 36^{\circ}+4\sin^2 30^{\circ}$
=

$\displaystyle 2\sin^2 (90 - 36)^{\circ}-2\cos ^2\ 36^{\circ}+4\sin^2 30^{\circ}$
=

$\displaystyle 2\cos^2 36^{\circ}-2\cos ^2\ 36^{\circ}+4\sin^2 30^{\circ}$
=

$4 \sin^2 30^{\circ}$
=

$4 (\dfrac{1}{2})^2$
=

$1$

Refer to the triangle given in figure.

Find: $|\tan ^{2}A\, -\, \sec ^{2}A|$

Given,

$\triangle ABC$ , A perpendicular from B on AC, let it cut AC at D, such that

$\angle BDA = \angle BDC = 90^{\circ}$

$AD = 3$

$BD = 4$

$BC = 12$

$\triangle ABD$ ,

$AB^2 = BD^2 + AD^2$

$AB^2 = 3^2 + 4^2$

$AB = 5$

$\triangle BCD$

$BC^2 = CD^2 + BD^2$

$12^2 = CD^2 + 4^2$

$CD^2 = 128$

$CD = 8 \sqrt{2}$

Now,

$|\tan^2 A - \sec^2 A| = \left|\left(\dfrac{P}{B}\right)^2 - \left(\dfrac{H}{B}\right)^2\right| = \left|\left(\dfrac{4}{3}\right)^2 - \left(\dfrac{5}{3}\right)^2\right|$

$\left|\dfrac{16}{9} - \dfrac{25}{9}\right|$ =

$\left|\dfrac{16 - 25}{9}\right|$

$\left|\dfrac{-9}{9}\right| = 1$

In an Isosceles triangle ABC, $tan^{2}\, B\, -\,sec^{2}\, B\, +\, 2$

$AB = AC = 15$ ,

$BC = 18$

Using cosine rule,

$AC^2 = BC^2 + AB^2 - 2 (AB)(BC) \cos B$

$15^2 = 18^2 + 15^2 - 2 (15) (18) \cos B$

$\cos B = \dfrac{9}{15}$

$\cos B = \dfrac{3}{5}$

$\cos B = \dfrac{B}{H} = \dfrac{3}{5}$

Now, using Pythagoras theorem.

$H^2 = P^2 + B^2$

$5^2 = P^2 + 3^2$

$P = 4$

$\tan^2 B - \sec^2B + 2$
=

$(\dfrac{P}{B})^2 - (\dfrac{H}{B})^2 + 2$
=

$(\dfrac{4}{3})^2 - (\dfrac{5}{3})^2 + 2$
=

$\dfrac{16 - 25 + 18}{9}$
=

$1$

If $cot\, \Theta\, =\, 1$ ; find the value of :
$5\, tan^{2} \Theta\, +\, 2\, sin^{2}\, \Theta\, -\, 3$

$\cot \Theta = 1$

$\cot \Theta = \cot 45$

$\Theta = 45^{\circ}$

Now,

$5\, tan^{2} \Theta\, +\, 2\, sin^{2}\, \Theta\, -\, 3$
=

$5\, tan^{2}45\, +\, 2\, sin^{2}\,45\, -\, 3$
=

$5 (1)^2 + 2 (\frac{1}{\sqrt{2}})^2 - 3$
=

$5 + 1 - 3$
=

$3$

$sin^{2}\, B\, +\, cos^{2}\, C$

Given,

$AB = AC = 10$ and

$BC = 18$ cm

$\cos B = \frac{AB^2 + BC^2 - AC^2}{2 AB \times BC}$

$\cos B = \frac{10^2 + 18^2 - 10^2}{2\times10 \times 18}$

$\cos B = \frac{9}{10}$

$\cos B = \frac{B}{H} = \frac{9}{10}$
Using Pythagoras Theorem,

$H^2 = P^2 + B^2$

$10^2 = P^2 + 9^2$

$P = \sqrt{19}$

$\sin B = \frac{P}{H} = \frac{\sqrt{19}}{10}$

Now,

$\cos C =\frac{AC^2 + BC^2 - AB^2}{2 AC \times BC}$

$\cos C = \frac{10^2 + 18^2 - 10^2}{2\times10 \times 18}$

$\cos C = \frac{9}{10}$

Thus,

$\sin^2 B + \cos^2 C = (\frac{\sqrt{19}}{10})^2 + (\frac{9}{10})^2$

$\sin^2 B + \cos^2 C= \frac{19}{100} + \frac{81}{100}$

$\sin^2 B + \cos^2 C= \frac{100}{100} = 1$

If sin A = cos A, find the value of $2\, tan^{2}\, A\, -\, 2\, sec^{2}\, A\, +\, 5$ .

Given,

$sin A = cos A$

$tan A = 1$

$tan A = tan 45$

$A = 45^{\circ}$

To Prove:

$2\, tan^{2}\, A\, -\, 2\, sec^{2}\, A\, +\, 5$
=

$2\, tan^{2}\, 45\, -\, 2\, sec^{2}\, 45\, +\, 5$
=

$2 (1)^2 - 2(\sqrt{2})^2 + 5$
=

$2 - 4 + 5$
=

$3$

$3\, -\, cot^{2}\, A\, +\, cosec^{2}\, A$

$3 \cos A = 4 \sin A$

$\tan A = \dfrac{3}{4}$

$\tan A = \dfrac{P}{B} = \dfrac{3}{4}$

Now, Using Pythagoras Theorem,

$H^2 = P^2 + B^2$

$H^2 = 3^2 + 4^2$

$H = 5$

Now,

$3 - \cot^2 A + cosec^2 A = 3 - (\dfrac{B}{P})^2 + (\dfrac{H}{P})^2$
=

$3 - (\dfrac{4}{3})^2 + (\dfrac{5}{3})^2$
=

$3 - \dfrac{16}{9} + \dfrac{25}{9}$
=

$\dfrac{27 - 16 + 25}{9}$
=

$4$

$(sec\, x^{\circ}\, -\, tan\, x^{\circ})\, (sec\, x^{\circ}\, +\, tan\, x^{\circ})$

Let the given triangle be ABC, where

$\angle ABC = 90^{\circ}$ ,

$AB = 1, BC = y, AC = 2$

By Pythagoras Theorem,

$AC^2 = AB^2 + BC^2$

$2^2 = 1^2 + y^2$

$y^2 = 3$

$y = \sqrt{3}$

Now,

$\left ( \sec x^{0}\, -\, \tan x^{0} \right )\left ( \sec x^{0}\, +\, \tan x^{0} \right )$
=

$\sec^2 x^{\circ} - \tan^2 x^{\circ}$
=

$(\dfrac{H}{B})^2 - (\dfrac{P}{B})^2$
=

$(\dfrac{2}{1})^2 - (\dfrac{\sqrt{3}}{1})^2$
=

$4 - 3$
=

$1$

$tan^{2}\, C\, -\, sec^{2}\, B\, +\, 2$

Given,

$AB = AC = 10$ and

$BC = 18$ cm

$\cos B = \dfrac{AB^2 + BC^2 - AC^2}{2 AB \times BC}$

$\cos B = \dfrac{10^2 + 18^2 - 10^2}{2\times10 \times 18}$

$\cos B = \dfrac{9}{10}$

$\cos B = \dfrac{B}{H} = \dfrac{9}{10}$

Now,

$\cos C =\dfrac{AC^2 + BC^2 - AB^2}{2 AC \times BC}$

$\cos C = \dfrac{10^2 + 18^2 - 10^2}{2\times10 \times 18}$

$\cos C = \dfrac{9}{10}$
Using Pythagoras Theorem,

$H^2 = P^2 + B^2$

$10^2 = P^2 + 9^2$

$P = \sqrt{19}$

$\tan C = \dfrac{P}{B} = \dfrac{\sqrt{19}}{9}$

Thus,

$\tan^2 C - \sec^2 B + 2 = (\dfrac{\sqrt{19}}{9})^2 - (\dfrac{10}{9})^2 + 2$

$\tan^2 C - \sec^2 B= \dfrac{19}{81} - \dfrac{100}{81} + 2$

$\tan^2 C - \sec^2 C + 2= \dfrac{-81}{81} + 2= 1$

Find the value of:
$\text{cosec}^{2}\,45^{\circ}\,-\,\cot^{2}\,45^{\circ}\,$

$\text{cosec}^{2}\,45^{\circ}\,-\,\cot^{2}\,45^{\circ}$
=

$(\sqrt{2})^2 - 1^2$
=

$2 - 1$
=

$1$

$\cos^{2}\, 42^{\circ}\, -\, \sin^{2}\, 48^{\circ}$

$\cos^{2}\, 42^{\circ}\, -\, \sin^{2}\, 48^{\circ}$
=

$\cos^{2}\, (90 - 48)^{\circ}\, -\, \sin^{2}\, 48^{\circ}$
=

$\sin^{2}\, 48^{\circ}\, -\, \sin^{2}\, 48^{\circ}$
=

$0$

$cosec^{2}\, 60^{\circ}\, -\, \tan^{2}\, 30^{\circ}$

$\csc^2 60^{\circ} - \tan^2 30^{\circ}$
=

$(\dfrac{2}{\sqrt{3}})^2 - (\dfrac{1}{\sqrt{3}})^2$
=

$\dfrac{4}{3} - \dfrac{1}{3}$
=

$\dfrac{3}{3}$
=

$1$

If $\displaystyle \sin \theta +\cos \theta =1$ , then the value of $\displaystyle \sin \theta \cos \theta$ is

$\displaystyle \sin \theta +\cos \theta =1$ or squaring both sides

$\displaystyle \sin ^{2}\theta +\cos ^{2}\theta +2\sin \theta \cos \theta =1$
i.e.

$\displaystyle 1+2\sin \theta \cos \theta =1$
or

$\displaystyle 2\sin \theta \cos \theta =0$ or

$\displaystyle \sin \theta \cos \theta =0$

If $\displaystyle \sec \theta =x+\frac{1}{4x}$ , then the value of $\displaystyle \sec \theta +\tan \theta$ is equal to

$\displaystyle \tan ^{2}\theta =\sec ^{2}\theta -1$

$\displaystyle =\left ( x+\frac{1}{4x} \right )^{2}-1$

$\displaystyle =\left ( x+\frac{1}{4x} \right )^{2}$

$\displaystyle \therefore \tan \theta =\pm \left ( x-\frac{1}{4x} \right )$

$\displaystyle =\left ( x-\frac{1}{4x} \right )or\left ( \frac{1}{4x}-x \right )$

$\displaystyle \therefore \sec \theta +\tan \theta$ is either 2x or

$\displaystyle \frac{1}{2x}$

If $\displaystyle A=29^{0}$ , then find the value of $\displaystyle \sin ^{4}A+\cos ^{4}+A+2\sin ^{2}A\cos ^{2}A$ .

since

$\displaystyle \sin ^{4}A+\cos ^{4}A+2\sin ^{2}A\cos ^{2}A$

$\displaystyle =\left ( \sin ^{2}A+\cos ^{2}A \right )=1$ for all value of A
Hence the value of the given expression is 1

The value of $\displaystyle \left ( \sin x\cos y+\cos x\sin y \right )^{2}+\left ( \cos x\cos y-\sin x\sin y \right )^{2}$ is equal to

The given expression is equal to

$\displaystyle \sin ^{2}\left ( x+y \right )+\cos ^{2}\left ( x+y \right )=1$

What is the value of $\displaystyle \frac{\left ( \sec \theta +1 \right )^{2}+\left ( 1-\sec \theta \right )^{2}}{1+\sec ^{2}\theta }$ ?

$\displaystyle \frac{\left ( a+b \right )^{2}+\left ( a-b \right )^{2}}{a^{2}+b^{2}}=2$
In this case

$\displaystyle a=\sec \theta$ b=1 Hence the value of the given expression is 2

Prove that:
$\displaystyle 2\, cos\frac{\pi}{13}cos\frac{9\pi}{13}+cos\frac{3\pi}{13}+cos\frac{5\pi}{13}=0$

$LHS=\displaystyle2\, cos\frac { \pi }{ 13 } cos\frac { 9\pi }{ 13 } +cos\frac { 3\pi }{ 13 } +cos\frac { 5\pi }{ 13 }$

$\displaystyle =cos\frac { 10\pi }{ 13 } +cos\frac { 8\pi }{ 13 } +cos\frac { 3\pi }{ 13 } +cos\frac { 5\pi }{ 13 }$

$[\because 2 \cos A \cos B = \cos (A+B) + \cos (A-B)]$

$\displaystyle =\cos { \left( \pi -\frac { 3\pi }{ 13 } \right) } +\cos { \left( \pi -\frac { 5\pi }{ 13 } \right) } +cos\frac { 3\pi }{ 13 } +cos\frac { 5\pi }{ 13 }$

$[\because \cos(\pi-\theta)=-\cos \theta]$

$\displaystyle =-cos\frac { 3\pi }{ 13 } -cos\frac { 5\pi }{ 13 } +cos\frac { 3\pi }{ 13 } +cos\frac { 5\pi }{ 13 } =0=RHS$

Hence proved

Prove that:
$\displaystyle 2sin^2\frac{3\pi}{4}+2cos^2\frac{\pi}{4}+2sec^2\frac{\pi}{3}=10$

$LHS=2sin^{ 2 }\dfrac { 3\pi }{ 4 } +2cos^{ 2 }\dfrac { \pi }{ 4 } +2sec^{ 2 }\dfrac { \pi }{ 3 }$

$\displaystyle=2\sin ^{ 2 }{ \left( \pi -\frac { \pi }{ 4 } \right) } +2{ \left( \frac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }+2{ \left( 2 \right) }^{ 2 }=2{ \left( \frac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }+1+8=10=RHS$

Find the values of $\displaystyle \theta$ and $p$ , if the equation $\displaystyle x\cos \theta +y\sin \theta = p$ is the normal form of the line $\displaystyle \sqrt{3}x+y+2= 0.$

The equation of the given line is

$\displaystyle \sqrt{3}x+y+2= 0$
this equation can be reduced as

$\displaystyle -\sqrt{3}x-y= 2$
on dividing both sides by

$\displaystyle \sqrt{(-\sqrt{3})^2+(-1)^{2}}= 2,$
we obtain

$\displaystyle -\frac{\sqrt{3}}{2}x-\frac{1}{2}y= \frac{2}{2}$

$\displaystyle \Rightarrow \left \{ -\frac{\sqrt{3}}{2} \right \}x+\left \{ -\frac{1}{2} \right \}y= 1$ ....(1)
On comparing equation (1) to

$\displaystyle x\cos \theta +y\sin \theta = p$ ,
we obtain

$\displaystyle \cos \theta = -\frac{\sqrt{3}}{2},\sin \theta = -\frac{1}{2},$ and

$p= 1$
Since the value of

$\displaystyle \sin \theta$ and

$\cos \theta$ are both negative,

$\theta$ is in the third quadrant

$\displaystyle \therefore \theta = \pi +\frac{\pi }{6}= \frac{7\pi }{6}$

Thus, the respective values of

$\displaystyle \theta$ and

$p$ are

$\cfrac{7\pi }{6}$ and

$1$

Prove that:
$\displaystyle 2sin^2\frac{\pi}{6}+cosec^2\frac{7\pi}{6}cos^2\frac{\pi}{3}=\frac{3}{2}$

LHS

$=2sin^2\dfrac {\pi}{6}+cosec^2 \dfrac {7\pi}{6}cos^2\dfrac {\pi}{3}$

$=2sin^2\dfrac {\pi}{6}+cosec^2 \left(\pi+ \dfrac {\pi}{6}\right) cos^2\dfrac {\pi}{3}$

$=2sin^2\dfrac {\pi}{6}+cosec^2 \left(- \dfrac {\pi}{6}\right) cos^2\dfrac {\pi}{3}$ ....As

$\theta$ lies in the 3rd quadrant,

$\theta$ is negative.

$=2 \left(\dfrac 12\right)^2 + (-2)^2 \times \left(\dfrac 12 \right)^2$

$=2 \times \dfrac 14 + 1$

$= \dfrac 32$

$=$ RHS

Hence proved.

Prove the following:
$\displaystyle \cos \left(\dfrac{3\pi}{2}+x\right) \cos(2\pi+x)\left[\cot\left(\dfrac{3\pi}{2}-x\right)+\cot (2\pi +x)\right]=1$

$LHS\displaystyle=cos\left( \frac { 3\pi }{ 2 } +x \right) cos(2\pi +x)\left[ cot\left( \frac { 3\pi }{ 2 } -x \right) +cot(2\pi +x) \right]$

$=\sin { x } \cos { x } \left[ \tan { x } +\cot { x } \right]$

$\displaystyle =\frac { \sin { x } \cos { x } \left( \sin ^{ 2 }{ x } +\cos ^{ 2 }{ x } \right) }{ \sin { x } \cos { x } } =1=RHS$

Hence proved

(i) If A + B = $45^o$ , prove that $\displaystyle (cot A-1)(cotB-1)=2$ and hence deduce that $\displaystyle cot 22 \frac{1}{2}^o = \sqrt{2} + 1$ .
(ii) If $\displaystyle tan(A-B) = \frac{7}{24}$ and $\displaystyle tan A=\frac{4}{3}$ where A and B are acute, show that A + B = $\displaystyle \frac{\pi}{2}$

(i)

$cot(A+B)=\dfrac{cotA.cotB-1}{cotA+cotB}$

and

$A+B=45$

$cotA.cotB-1=cotA+cotB$

$cotA.cotB-cotA-cotB-1+2=2$

$(cotA-1)(cotB-1)=2$

$A=B= \dfrac{\pi}{8}$ , and

$cot\dfrac{\pi}{8}=x$

$(x-1)(x-1)=2$

$x^2-2x-1=0$

$x=\sqrt{2}+1$

(ii)

$tan(A-B)=\dfrac{tanA-tanB}{1+tanA.tanB}$

$\dfrac{7}{24}=\dfrac{\dfrac{4}{3}-tanB}{1+\dfrac{4}{3} tanB}$

=> simplifying, we get

$tanB=\dfrac{3}{4}$

Now,

$tan(A+B)= \dfrac{tanA+tanB}{1-tanA.tanB}$

$1-tanA.tanB= 1-\dfrac{4}{3}.\dfrac{3}{4}=0$

$A+B=\dfrac{\pi}{2}$

Express the trigonometric ratio tan A in terms of sec A.

Since,

$sec^2A-tan^2A=1$

$\Rightarrow tan^2A=sec^2A-1$

$\Rightarrow tanA=\pm \sqrt{sec^2A-1}$

(i) If $\cos 3A = \sin (A - 34^0)$ , where A is an acute angle, find the value of A.
(ii) Prove the following identity, where the angles involved are acute angles for which the expression is define.
$\dfrac{1 + \cot^2 A}{1 + \tan^2 A} = \left( \dfrac{1 - \cot A}{1 - \tan A} \right )^2$

(i)

$\cos { 3A } =\sin { \left( A-34 \right) } \quad$

$=\cos { (90-A+34) }$

$=\cos { (124-A) }$

So,

$124-A=2n\pi \pm 3A$

For

$+ve$

$124-A=2n\pi + 3A$

$4A=124-2n\pi$

$A$ is acute

$n=0$

$A=\dfrac { 124 }{ 4 } =31°$

For

$-ve$

$124-A=2n\pi - 3A$

$124+2A=2n\pi$

$2A=2n\pi-124$

$A$ is acute, no

$n$ exists

$A=31°$

(ii)

$\dfrac { 1+\cot ^{ 2 }{ A } }{ 1+\tan ^{ 2 }{ A } } =\dfrac { \csc ^{ 2 }{ A } }{ \sec ^{ 2 }{ A } } =\cot ^{ 2 }{ A }$

${ \left( \dfrac { 1-\cot { A } }{ 1-\tan { A } } \right) }^{ 2 }={ \left( \dfrac { 1-\dfrac { 1 }{ \tan { A } } }{ 1-\tan { A } } \right) }^{ 2 }={ \left( \dfrac { \tan { A } -1 }{ 1-\tan { A } } \right) }^{ 2 }\times \dfrac { 1 }{ \tan ^{ 2 }{ A } } =\cot ^{ 2 }{ A }$

So,

$\dfrac { 1+\cot ^{ 2 }{ A } }{ 1+\tan ^{ 2 }{ A } } ={ \left( \dfrac { 1-\cot { A } }{ 1-\tan { A } } \right) }^{ 2 }$

If $\tan \theta + \sin \theta = m, \tan \theta - \sin \theta = n$ and $m\neq n$ , then show that $m^{2} - n^{2} = 4\sqrt {mn}$

Given,

$m=\tan \theta+\sin \theta, n=\tan \theta -\sin \theta$

We need to show

$m^2-n^2=4\sqrt {mn}$

Taking L.H.S.,

$m^2-n^2=(\tan\theta+\sin \theta)^2-(\tan\theta -\sin\theta)^2$

$=\tan^2\theta+\sin^2\theta+2\tan \theta. \sin\theta-\tan^2\theta-\sin^2\theta+2\tan \theta. \sin\theta$

$=4\tan\theta. \sin\theta$

Now, taking R.H.S.,

$4\sqrt {mn}=4\sqrt {(\tan\theta+\sin\theta)(\tan\theta-\sin\theta)}$

$=4\sqrt {\tan^2\theta-\sin^2\theta}$

$=4\sqrt {\dfrac {\sin^2\theta}{\cos^2\theta}-\sin^2\theta}$

$=4\sqrt {\dfrac {\sin^2\theta(1-\cos^2\theta)}{\cos^2\theta}}$

$=4\sqrt {\sin^2\theta.\tan^2\theta}$

$=4\tan\theta. \sin\theta$

Therefore, L.H.S.

$=$ R.H.S.

Prove the following identity
$\sqrt { \sec ^{ 2 }{ \theta } +cosec ^{ 2 }{ \theta } } =\tan { \theta } +\cot { \theta }$

We need to prove

$\sqrt {\sec^2\theta +cosec ^2\theta}=\tan \theta +\cot \theta$

L.H.S.

$=\sqrt { \sec ^{ 2 }{ \theta } +cosec ^{ 2 }{ \theta } }$

$=\sqrt { 1+\tan ^{ 2 }{ \theta } +1+\cot ^{ 2 }{ \theta } }$

$=\sqrt { \tan ^{ 2 }{ \theta } +\cot ^{ 2 }{ \theta } +2 }$

$=\sqrt { \tan ^{ 2 }{ \theta } +\cot ^{ 2 }{ \theta } +2\tan { \theta } \cot { \theta } }$

$=\sqrt { { \left( \tan { \theta } +\cot { \theta } \right) }^{ 2 } }$

$=$ R.H.S.

If $3 \cot A = 4$ , then evaluate $\dfrac{1 - \tan^2 A}{1 + \tan^2 A}$

Given,

$3cotA=4$

$\Rightarrow tanA=\frac34$

$\Rightarrow tan^2A=\frac9{16}$

$\displaystyle \Rightarrow\frac{1-tan^2A}{1+tan^2A}=\frac{1-\frac9{16}}{1+\frac9{16}}=\frac{\frac7{16}}{\frac{25}{16}}=\frac7{25}$

Find the value of $\cfrac { \sec { { 15 }^{ o } } }{ co\sec { { 75 }^{ o } } } +\cfrac { \sin { { 72 }^{ o } } }{ \cos { { 18 }^{ o } } } -\cfrac { \tan { { 33 }^{ o } } }{ \cot { { 57 }^{ o } } }$ .

To find:

$\dfrac{sec15^\circ}{cosec75^\circ}+\dfrac{sin72^\circ}{cos18^\circ}-\dfrac{tan33^\circ}{cot57^\circ}=?$

Solution:

$\dfrac{sec15^\circ}{cosec75^\circ}+\dfrac{sin72^\circ}{cos18^\circ}-\dfrac{tan33^\circ}{cot57^\circ}$

$=\dfrac{sec(90^\circ-75^\circ)}{cosec75^\circ}+\dfrac{sin(90^\circ-18^\circ)}{cos18^\circ}-\dfrac{tan(90^\circ-57^\circ)}{cot57^\circ}$

$=\dfrac{cosec75^\circ}{cosec75^\circ}+\dfrac{cos18^\circ}{cos18^\circ}-\dfrac{cot57^\circ}{cot57^\circ}$

$(\because sin(90^\circ-\theta)=cos\theta,$

$sec(90^\circ-\theta)=cosec\theta,$

$tan(90^\circ-\theta)=cot\theta),$

$=1+1-1$

$=2-1$

$=1$

Prove: $(\sin^{6}\theta + \cos^{6}\theta) = 1 - 3\sin^{2}\theta \cos^{2}\theta$

We need to prove

$(\sin^6\theta+\cos ^6\theta)=1-3\sin^2\theta.\cos^2\theta$

Taking L.H.S.

$= \sin^{6} \theta + \cos^{6}\theta$

$= (\sin^{2}\theta) + (\cos^{2}\theta)^{3}$

$= (\sin^{2}\theta + \cos^{2}\theta) (\sin^{4}\theta - \sin^{2}\theta \cos^{2}\theta + \cos^{4}\theta)$

$= 1(\sin^{4}\theta - \sin^{2}\theta \cos^{2}\theta + \cos^{4}\theta)$

$= \sin^{4}\theta - \sin^{2}\theta \cos^{2}\theta + \cos^{4}\theta$

$= (\sin^{2}\theta)^{2} + (\cos^{2}\theta)^{2} - \sin^{2}\theta \cos^{2}\theta$

$[\therefore a^{2} + b^{2} = (a + b)^{2} - 2ab]$

$= (\sin^{2}\theta + \cos^{2}\theta)^{2} - 2\sin \theta \cos^{2}\theta - \sin^{2}\theta \cos^{2}\theta$

$= 1 - 3\sin^{2}\theta \cos^{2}\theta$

$=$ R.H.S.

Prove that $\sqrt { \cfrac { 1-\sin { \theta } }{ 1+\sin { \theta } } } =\sec { \theta } -\tan { \theta }$ , (where $\theta$ is acute).

Given LHS ,

$\sqrt { \dfrac { 1-\sin { \theta } }{ 1+\sin { \theta } } }$

Multiplying numerator and denominator under root with

$1-\sin{\theta}$

We get

$\dfrac{1-\sin{\theta}}{\cos{\theta}}$ which will be equal to

$\sec{\theta} - \tan{\theta}$

Prove that ${ \left( \sin { \theta } +co\sec { \theta } \right) }^{ 2 }+{ \left( \cos { \theta } +\sec { \theta } \right) }^{ 2 }=7+\tan ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta }$

We need to prove

${ \left( \sin { \theta } +co\sec { \theta } \right) }^{ 2 }+{ \left( \cos { \theta } +\sec { \theta } \right) }^{ 2 }=7+\tan ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta }$
L.H.S.

$={ \left( \sin { \theta } +co\sec { \theta } \right) }^{ 2 }+{ \left( \cos { \theta } +\sec { \theta } \right) }^{ 2 }$

$=\sin ^{ 2 }{ \theta } +co\sec ^{ 2 }{ \theta } +2\sin { \theta } co\sec { \theta } +\cos ^{ 2 }{ \theta } +\sec ^{ 2 }{ \theta } +2\cos { \theta } \sec { \theta }$

$=\sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } +2+2+co\sec ^{ 2 }{ \theta } +\sec ^{ 2 }{ \theta }$

$=1+2+2+1+\cot ^{ 2 }{ \theta } +1+\tan ^{ 2 }{ \theta }$

$=7+\tan ^{ 2 }{ \theta } +\cot ^{ 2 }{ \theta }$

$=$ R.H.S.

Show that $\tan ^{ 2 }{ \theta } -\cfrac { 1 }{ \cos ^{ 2 }{ \theta } } =-1$ .

Formula used:

$\displaystyle sec^2\theta-tan^2\theta=1$

$\displaystyle tan^2\theta-\frac1{cos^2\theta}=tan^2\theta-sec^2\theta$

$=-(sec^2\theta-tan^2\theta)$

$=-1$

If $(2\cos x + \sin x) = 1$ , then sum of all possible value of $(7\cos x+ 6 \sin x)$ is__________

Given

$2\cos x+\sin x=1\implies \dfrac{2(1-\tan^2 \frac{x}{2})}{1+\tan^2 \frac{x}{2}}+\dfrac{2\tan \frac{x}{2}}{1+\tan^2 \frac{x}{2}}=1$

$\implies 2-2\tan^2 \frac{x}{2}+2\tan \frac{x}{2}=1+\tan^2 \frac{x}{2}$

$\implies 3\tan^2 \frac{x}{2}-2\tan \frac{x}{2}-1=0$

$\implies 3\tan^2 \frac{x}{2}-3\tan \frac{x}{2}+\tan \frac{x}{2}-1=0$

$\implies (\tan \frac{x}{2}-1)(3\tan \frac{x}{2}+1)=0$

$\tan \frac{x}{2}=1,-\dfrac{1}{3}$

When

$\tan \dfrac{x}{2}=1$

then

$7\cos x+6\sin x=7\bigg(\dfrac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}\bigg)+6\bigg(\dfrac{2\tan \frac{x}{2}}{1+\tan^2 \frac{x}{2}}\bigg)=7\times \dfrac{1-1}{1+1}+6\times \dfrac{2\times 1}{1+1}=0+6=6$

When

$\tan \dfrac{x}{2}=-\dfrac{1}{3}$

then

The sum of all possible values is

$6+2=8$

Evaluate: $\dfrac {1}{\sec A - \tan A} - \dfrac {1}{\cos A} = \dfrac {1}{\cos A} - \dfrac {1}{(\sec A + \tan A)}$ .

Taking LHS=

$\dfrac{1}{\sec A-\tan A}-\dfrac{1}{\cos A}=\dfrac{1}{\sec A - \tan A}-\sec A$

$=\dfrac{1-(\sec A)^2 +\sec A \times \tan A}{\sec A -\tan A}$

Multiply Numerator and Denominator by

$(\sec A + \tan A)$

$=\dfrac{(1-(\sec A)^2 +\sec A \times \tan A) \times (\sec A +\tan A)}{(\sec A -\tan A)\times (\sec A + \tan A)}$

We know

$(\sec A)^2-(\tan A)^2=1$

Thus we have

$\dfrac{(-(\tan A)^2 +\sec A \times \tan A)(\sec A +\tan A)}{(\sec A -\tan A)(\sec A + \tan A)}$

$=\dfrac{(\tan A)(\sec A +\tan A)}{(\sec A + \tan A)}$

Again using the above identity,

$=\dfrac{(\sec A)^2 -1 +\sec A \times \tan A}{\sec A +\tan A}$

$=\sec A -\dfrac{1}{\sec A +\tan A}=$ RHS

Prove by vector method that
$\sin { \left( \alpha +\beta \right) } =\sin { \alpha } \cos { \beta } +\cos { \alpha } \sin { \beta }$

Let

$OQ=OP=1$

$OQ$ makes angle

$\alpha$ and

$OP$ makes angle

$\beta$

with x-axis

Let

$PM$ and

$QN$ be perpendicular to x-axis

$\overrightarrow { OQ }=\overrightarrow {ON } +\overrightarrow { NQ }$

$=|ON|\hat{i}+|NQ|\hat{j}$

$=(OQ\cos \alpha)\hat{i}+(OQ\sin\alpha)\hat{j}$

$[\because OQ=1]$

$=\cos\alpha\hat{i}+\sin\alpha\hat{j}$ ........(1)

Similarly

$\overrightarrow { OP } =\overrightarrow { OM } +\overrightarrow { MP }$

$=|OM|\hat{i}+|MP|\hat{j}$

$=(OP\cos\beta)\hat{i}+(OP\sin \beta)-\hat{j}$

$[\because OP=1]$

$=\cos\beta\hat{i}-\sin\beta\hat{j}$ .......(2)

$\overrightarrow { OP} \times \overrightarrow { OQ } =(\cos\beta\hat{i}-\sin\beta\hat{j})\times (\cos\alpha\hat{i}\sin\alpha\hat{j})$

$\cos\beta\cos\alpha(\hat{i}\times \hat{i})+\cos\beta\sin\alpha(\hat{i}\times \hat{j})-\sin\beta\cos\alpha(\hat{j}\times \hat{i})-\sin\beta\sin\alpha(\hat{j}\times \hat{j})$

$=(\sin\alpha\cos\beta+\cos\alpha\sin\beta)\hat{k}$ .....(i)

$\overrightarrow { OP } \times \overrightarrow {OQ }=|OP||OQ|\sin(\alpha+\beta)\hat{k}$

$=\sin(\alpha+\beta)\hat{k}$ ......(iii)

By (i) and (ii)

$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$

1450477_776382_ans_bf9d6d0dfe894d368a36f484ee3d8014.png

Prove that in a triangle with angles A,B,C and opposite sides as a,b,c $\frac{\sin{\left(B-C\right)}}{\sin{\left(B+C\right)}} = \frac{{b}^{2}-{c}^{2}}{{a}^{2}}$

To prove

$\to \ \dfrac {\sin (B-C)}{\sin (B+C)}\dfrac {b^2 -c^2}{a^2}$

where

$A, B, C, \to$ angles of

$\triangle$

$a, b, c\to$ sides of

$\triangle$

defined such that,

where

$A+B+C=\pi$

Using Sine Rule for a

$\triangle$

$\dfrac {\sin A}{a}=\dfrac {\sin B}{b}=\dfrac {\sin C}{c}=b (let)$

From here, we get

$a=\dfrac {\sin A}{b}, b=\dfrac {\sin B}{b}, c=\dfrac {\sin C}{b}$

$\Rightarrow \$ Put values

$a, b, c$ on

$RHS$

$RHS=\dfrac {b^2 -c^2}{a^2}=\dfrac {\dfrac {\sin^2 B}{b^2}-\dfrac {\sin^2 C}{b^2}}{\dfrac {\sin^2 A}{b^2}}$

$=\dfrac {\sin^2 B-\sin^2 C}{\sin^2 A}$

$[Using\ Formula \to \sin (B+C) \sin (B-C)=\sin^2 B-\sin^2C]$

$=\dfrac {\sin (B+C) \sin (B-C)}{\sin^2 A}\quad [As\ A+B+C=\pi \\ A=\pi -(B+C)]$

$=\dfrac {\sin (B+C) \sin (B-C)}{\sin^2 (\pi -(B+C))}$

$=\dfrac {\sin (B-C)}{\sin^2 (B+C)}$

$=\dfrac {\sin (B-C)}{\sin (B+C)}=LHS$

Hence proved.

1443141_879793_ans_4d89f0ec9e4444d18fdf39c2cd778660.png

If $f_n(\alpha)=\displaystyle\frac{\sin\alpha +\sin 3\alpha +\sin 5\alpha +.....+\sin(2n-1)\alpha}{\cos\alpha +\cos 3\alpha +\cos 5\alpha + .....+\cos(2n-1)\alpha}$ find $f_4(\pi /32)$ .

Using,

$\displaystyle \sum_{k=0}^{n-1}{\cos (a+kd)}=\dfrac{\sin \left(n\times d/2\right)}{\sin \dfrac{d}{2}}\times \cos \left(\dfrac{2a+(n-1)d}{2}\right)$

$\displaystyle \sum_{k=0}^{n-1}{\sin (a+kd)}=\dfrac{\sin \left(n d/2\right)}{\sin \dfrac{d}{2}}\times \sin \left(\dfrac{2a+(n-1)d}{2}\right)$

$f_n(\alpha)=\dfrac{\sin \alpha +\sin 3\alpha +.....+\sin (2n-1)\alpha}{\cos \alpha +\cos 3\alpha +....+\cos (2x-1)\alpha}$

Here,

$a=\alpha$ &

$d=2\alpha$

$\therefore f_n(\alpha)=\dfrac{\dfrac{\sin \left(n d/2\right)}{\sin \dfrac{d}{2}}\times \sin \left(\dfrac{2a+(n-1)d}{2}\right)}{\dfrac{\sin \left(n\times d/2\right)}{\sin \dfrac{d}{2}}\times \cos \left(\dfrac{2a+(n-1)d}{2}\right)}$

$=\dfrac{\sin (\alpha +(n-1)\alpha }{\cos (\alpha +(n-1)\alpha}$ [ As

$\dfrac{\sin \left(n d/2\right)}{\sin (d/2)}\neq 0$

$=\tan (n\alpha)$

$\therefore f_4(32)=\tan \left(4.\dfrac{\pi}{32}\right)=\tan \dfrac{\pi}{8}=\sqrt 2-1$

If $\csc { \theta -\sin { \theta ={ a }^{ 3 } } }$ and $\sec { \theta } -\cos { \theta } =b^{ 3 }$ , prove that ${ a }^{ 2 }{ b }^{ 2 }\left( { a }^{ 2 }+{ b }^{ 2 } \right) =1$

${ a }^{ 3 }=\cfrac { { \cos }^{ 2 }\theta }{ \sin\theta } ,$

${ b }^{ 3 }=\cfrac { { \sin }^{ 2 }\theta }{ \cos\theta }$

${ a }^{ 2 }{ b }^{ 2 }({ a }^{ 2 }+{ b }^{ 2 })={ (\cfrac { { \cos }^{ 2 }\theta }{ \sin\theta } ) }^{ \frac { 2 }{ 3 } }{ (\cfrac { { \sin }^{ 2 }\theta }{ \cos\theta } ) }^{ \frac { 2 }{ 3 } }[{ (\cfrac { { \cos }^{ 2 }\theta }{ \sin\theta } ) }^{ \frac { 2 }{ 3 } }+{ (\cfrac { \sin^{ 2 }\theta }{ \cos\theta } ) }^{ \frac { 2 }{ 3 } }]$

${ a }^{ 2 }{ b }^{ 2 }({ a }^{ 2 }+{ b }^{ 2 })={ (\sin\theta \cos\theta ) }^{ \frac { 2 }{ 3 } }\times \cfrac { ({ \cos }^{ 2 }\theta +{ \sin }^{ 2 }\theta ) }{ { (\cos\theta \sin\theta ) }^{ \frac { 2 }{ 3 } } }$

${ a }^{ 2 }{ b }^{ 2 }({ a }^{ 2 }+{ b }^{ 2 })=1$

$(Proved)$

Prove that : $\dfrac { 1 }{ 1+\sin { \theta } } +\dfrac { 1 }{ 1-\sin { \theta } } =2\sec ^{ 2 }{ \theta }$

L.H.S

$= \cfrac{1}{1+\sin \theta}+\cfrac{1}{1-\sin \theta}$

$=\cfrac{1-\sin \theta+1+\sin \theta}{\left(1+\sin \theta\right)\left(1-\sin \theta\right)}$

$= \cfrac{2}{1-\sin^2\theta}=\cfrac{2}{\cos^2 \theta}=2\sec^2\theta=$ R.H.S

Prove that $\dfrac{\sin x - \sin 3x}{\sin^{2} x - \cos^{2} x} = 2 \sin x$

$\cfrac {sin x-\sin 3x }{ { \sin }^{ 2 }x-{ \cos }^{ 2 }x } \Longrightarrow \cfrac { \sin x-3\sin x+4{ \sin }^{ 3 }x }{ 2{ \sin }^{ 2 }x-1 }$

$\cfrac { 2\sin x(2{ \sin }^{ 2 }x-1) }{ 2{ \sin }^{ 2 }x-1 } =2\sin x$

$(Proved)$

If $\sec { \theta } =x+\dfrac { 1 }{ 4x }$ , then show that $\tan { \theta } +\sec { \theta } =2x$ or $\dfrac { 1 }{ 2x }$ .

Given,

$\sec \theta=x+\cfrac{1}{4x}$

So,

$\tan\theta=\sqrt{\sec ^2\theta-1}=\sqrt{\left(x+\cfrac{1}{4x}\right)^{2}-1}$

$=\sqrt{x^2+\cfrac{1}{16x^2}+\cfrac{1}{2}-1}=\sqrt{x^2+\cfrac{1}{16x^2}-\cfrac{1}{2}}$

$=\sqrt{x-\left(\cfrac{1}{4x}\right)^{2}}$

$\therefore \tan \theta=\pm \left(x-\cfrac{1}{4x}\right)$

Now,

$\sec \theta+\tan \theta= x+\cfrac{1}{4x}+\left(x-\cfrac{1}{4x}\right)=2x$

And,

$\sec \theta +\tan \theta =x+\cfrac { 1 }{ 4x } -\left( x-\cfrac { 1 }{ 4x } \right) =\cfrac{ 1} { 2x}$

Show that $sin^6 \theta + cos^6 \theta + 3 sin^2 \theta cos^2 \theta = 1$

L.H.S

$=\sin^6\theta+\cos^6\theta+3\sin^2\theta\cos^2\theta$

$=(\sin^2\theta)^3+(\cos^2\theta)^3+3\sin^2\theta\cos^2\theta$

$=(\sin^4\theta+\cos^4\theta)+(\sin^4\theta+\cos^4\theta-\sin^2\theta\cos^2\theta)+3\sin^2\theta\cos^2\theta$

$=\sin^$\theta+\cos^4\theta-\sin^2\theta\cos^2\theta+3\sin^2\theta\cos^2\theta$

$=\sin^4\theta+\cos^4\theta+2\sin^2\theta\cos^2\theta$

$=(\sin^2\theta+\cos ^2\theta)^2=1=$ R.H.S

Prove $\displaystyle \frac{\tan \, 2x \, \tan \, x}{\tan \, 2x \, - \, \tan \, x} \, = \, \sin \, 2x.$

$\cfrac{\tan 2x\tan x}{\tan 2x- \tan x}=\cfrac{\cfrac{2\tan x\tan x}{1-\tan^2 x}}{\cfrac{2\tan x}{1-\tan^2 x}-\tan x}$

$=\cfrac{2\tan^2 x}{2\tan x-\tan x\left(1-\tan^2 x\right)}=\cfrac{2\tan^2 x}{2\tan x-\tan x +\tan^3 x}$

$=\cfrac{2\tan x}{1+\tan^2x}=\cfrac{2\cfrac{\sin x}{\cos x}}{1+\cfrac{\sin^2 x}{\cos^2 x}}=\cfrac{\cfrac{2\sin x}{\cos x}}{\cfrac{\cos^2x+\sin^2x}{\cos^2x}}$

$=\cfrac{2\sin x\cos^2x}{\cos x}=2\sin x\cos x=\sin 2x=$ R.H.S

Prove the following identity:
$\displaystyle \sin^2 \, (45^{\circ} \, + \, \alpha) \, - \, \sin^2 \, (30^{\circ} \, - \, \alpha) \, - \, \sin \, 15^{\circ} \, \cos \, (15^{\circ} \, + \, 2\alpha) \, = \, \sin \, 2\alpha$

${ \sin }^{ 2 }A-{ \sin }^{ 2 }B=\sin(A+B)\sin(A-B)$

$=\sin(75°)\sin(15+2\alpha)-\sin 15°\cos(15+2\alpha)$

$=\cos(15°)\sin(15+2\alpha)-\sin 15°\cos(15+2\alpha)$

$=\sin(15+2\alpha-15)$

$=\sin 2\alpha$

Prove the following identity:
$\displaystyle \frac{1 \, - \, 2 \, \sin^2 \, \alpha}{1 \, + \, \sin \, 2\alpha} \, = \, \frac{1 \, - \, \tan \, \alpha}{1 \, + \, \tan \, \alpha}$

$\cfrac { 1-2{ \sin }^{ 2 }\alpha }{ 1+\sin2\alpha }$

$=\cfrac { \cos2\alpha }{ { (\sin\alpha +\cos\alpha ) }^{ 2 } }$

$=\cfrac { { \cos }^{ 2 }\alpha -{ \sin }^{ 2 }\alpha }{ { (\cos\alpha +\sin\alpha ) }^{ 2 } }$

$=\cfrac{\cos\theta-\sin \theta}{\cos\theta +\sin\theta}$

$=\cfrac{1-\tan \theta}{1+\tan \theta}$

Prove $\displaystyle \cos^2 \, \alpha \, - \, \sin^2 \, 2\alpha \, = \, \cos^2 \, \alpha \, \cos \, 2\alpha \, - \, 2 \, \sin^2 \, \alpha \, \cos^2 \, \alpha.$

L.H.S

$=\cos^2\alpha-\sin^2\alpha$

$=\cos^2\alpha-4\sin^2\alpha\cos^2\alpha\quad\left[\because \sin2\alpha=2\sin\alpha\cos\alpha\right]$

$=\cos^2\alpha-2\sin^2\alpha\cos^2\alpha-2\sin^2\alpha\cos^2\alpha$

$=\cos^2\alpha\left[1-2\sin^2\alpha\right]-2\sin^2\alpha\cos^2\alpha$

$=\cos^2\alpha\cos2\alpha-2\sin^2\alpha\cos^2\alpha$

R.H.S

Simplify the following expression:
$\displaystyle \frac{1 \, + \, \sin + \, 2x}{(\sin \, x \, + \, \cos \, x)^2}$

$=\cfrac{1+\sin2x}{\left(\sin x+\cos x\right)^2}=\cfrac{sin^2x+\cos^2x+\sin 2x}{\left(\sin x+\cos x\right)^2}$

$\left[\because 1=\sin^2x+cos^2x\right]$

$=\cfrac{\sin^2x+\cos^2x+2\sin x\cos x}{\left(\sin x+\cos x\right)^2}=\cfrac{\left( \sin x+\cos x \right) ^{ 2 }}{\left( \sin x+\cos x \right) ^{ 2 }}\quad \left[\because (a+b)^2=a^2+b^2+2ab\right]$

$=1$

Prove $\displaystyle 3 \, - \, 4 \, \cos \, 2\alpha \, + \, \cos \, 4\alpha \, = \, 8 \, \sin^4 \, \alpha.$

$3-4\cos 2\alpha+\cos 4 \alpha$

$=3-4(1-{2\sin}^{2}\alpha)+(1-{2\sin}^{2}2\alpha)$

$=3-4+{8\sin}^{2}\alpha +1-{8\sin}^{2}\alpha{\cos}^{2}\alpha$

$={8\sin}^{2}\alpha(1-{\cos}^{2}\alpha)$

$={8\sin}^{4}\alpha$

Prove the following identities.
$\displaystyle \frac{1 \, + \, \sin \, \alpha}{1 \, + \, \cos \, \alpha} \, .\frac{1 \, + \, \sec \, \alpha}{1 \, + \, \text{cosec} \, \alpha} \, = \, \tan \, \alpha.$

$LHS=\dfrac{1+\sin \alpha}{1+\cos \alpha}\ast \dfrac{1+\text{sec}\alpha}{1+\text{cosec}\alpha}$

$=\dfrac{1+\sin \alpha}{1+\cos \alpha}.\dfrac{1+\dfrac{1}{\cos\alpha}}{1+\dfrac{1}{\sin\alpha}}=\dfrac{\sin \alpha}{\cos\alpha}=\tan \alpha=RHS$

Solve the following:
$\displaystyle1 + \sec\, 20^{\circ} - \sqrt{3} \cot\, 40^{\circ}$

$1+\dfrac { 1 }{ \cos { 20 } } -\sqrt { 3 } \dfrac { \cos40 }{ \sin40 }$

$\\ \Longrightarrow \dfrac { \sin40-\sqrt { 3 } \cos40 }{ \sin40 } +\dfrac { 1 }{ \cos { 20 } }$

$\Longrightarrow 2\left( \dfrac { \frac { 1 }{ 2 } \sin40-\dfrac { \sqrt { 3 } }{ 2 } cos40 }{ \sin40 } \right) +\frac { 1 }{ \cos20 }$

$\Longrightarrow 2\left( \dfrac { -\sin20 }{ 2\sin20\cos20 } \right) +\dfrac { 1 }{ \cos20 }$

$\Longrightarrow \dfrac { -1 }{ \cos20 } +\dfrac { 1 }{ \cos20 } \\ =0$

Prove $\displaystyle \frac{1}{4 \, \sin^2 \, \alpha \, \cos^2 \, \alpha} \, - \, \frac{(1 \, - \, \tan^2 \, \alpha)^2}{4 \, \tan^2 \, \alpha} \, = \, 1.$

$\cfrac{1}{4\sin^2\alpha \cos^2\alpha}-\cfrac{\left(1-\tan^2\alpha\right)^{2}}{4\tan^2\alpha}$

$=\cfrac{1}{\left(\sin 2\alpha\right)^{2}}-\cfrac{\left[\cfrac{\cos^2\alpha- \sin^2\alpha}{\cos^2\alpha}\right]^2}{4\cfrac{\sin^2\alpha}{\cos^2\alpha}}$

$= \cfrac{1}{\left(\sin2\alpha\right)^{2}}-\cfrac{\cfrac{1}{4}\cfrac{\left(\cos 2\alpha\right)^{2}}{\cos^2\alpha \cos^2\alpha}\cos^2\alpha}{\sin^2\alpha}$

$=\cfrac{1}{\left(\sin2\alpha\right)^2}-\cfrac{\left(\cos 2\alpha\right)^{2}}{4\cos^2\alpha \sin^2\alpha}$

$=\cfrac{1-\left(\cos 2\alpha\right)^{2}}{\left(\sin 2\alpha\right)^{2}}=\cfrac{\left(\sin 2\alpha\right)^{2}}{\left(\sin 2\alpha\right)^{2}}=1$

Prove $\displaystyle \cot \, \alpha \, - \, \tan \, \alpha \, - \, 2 \, \tan \, 2\alpha \, - \, 4 \, \tan \, 4\alpha \, = 8 \, \cot 8 \, \alpha.$

$\cot \theta -\tan \theta=\cfrac{\cos 2\theta}{\sin \theta\cos\theta}=\cfrac{2\cos 2\theta}{\sin 2\theta}$

$\Longrightarrow \cot \alpha-\tan \alpha -2\tan 2\alpha=2\times \cfrac{\cos 2 \alpha}{\sin 2\alpha}-\cfrac{{2\sin} 2\alpha}{\cos 2 \alpha},$

$\Longrightarrow \cot \alpha-\tan \alpha -2\tan 2\alpha=2\times \cfrac{\cos 4 \alpha}{\sin 2\alpha\times{\cos} 2\alpha}=\cfrac{4\cos 4 \alpha}{\sin 4 \alpha}$

$\Longrightarrow \cot \alpha-\tan \alpha -2\tan 2\alpha-4\tan 4\alpha= \cfrac{4\cos 4 \alpha}{\sin 4}-\cfrac{4\sin 4 \alpha}{\cos 4 \alpha}$

$\Longrightarrow \cot \alpha-\tan \alpha -2\tan 2\alpha-4\tan 4\alpha= \cfrac{4\cos 8 \alpha}{\sin 4\cos 4\alpha}=\cfrac{8\cos 8 \alpha}{\sin 8 \alpha}$

$\Longrightarrow \cot \alpha-\tan \alpha -2\tan 2\alpha-4\tan 4\alpha= 8\cot 8\alpha$

Solve the following equation:
$\displaystyle \sin \, x \, - \, \sin \, 2x \,+ \, \sin \, 5x \, + \, \sin \, 8x \, = \, 0.$

$\sin x-\sin 2x+\sin 5x+sin 8x=0$

$2\sin 3x\cos 2x+2\cos 5x\sin 3x=0$

$2\sin 3x(\cos 2x+\cos 5x)=0$

$\sin 3x\cos\cfrac{7x}{2}\cos \cfrac{3x}{2}=0$

So,

$\sin 3x=0,$

$\cos \cfrac{7x}{2}=0$ and

$\cos\cfrac{3x}{2}=0.$

$\therefore$

$x=\cfrac { nx }{ 3 } ,x=(2n+1)\cfrac { \pi }{ 7 } ,x=(2n+1)\cfrac { \pi }{ 3 }$

Solve the following equation:
$\displaystyle (2 \, \sin \, x \, - \, \cos \, x) \, (1 \, + \, \cos \, x) \, = \, \sin^2 \, x.$

$(2\sin x-\cos x)(1+\cos x)={\sin}^{2} x$

$2\sin x+2\sin x\cos x-\cos x-{\cos}^{2}x={\sin}^{2}x$

$2\sin x+2\sin x\cos x-\cos x-1=0$

$2\sin x(1+\cos x)-1(1+\cos x)=0$

$(2\sin x-1)(1+\cos x)=0$

Either,

$(1+\cos x)=0\\ \cos x=-1,\\ x=(2n+1)\pi.$

Or,

$2\sin x-1=0,\\ \sin x=\cfrac{1}{2},\\ x=n\pi+{(-1)}^{n}\cfrac{\pi}{6}$

Solve the following equations.
$\displaystyle 4\sin^4 \, x \, + \cos^4 \, x=1$

${ 4\sin }^{ 4 }x+{ \cos }^{ 4 }x=1$

${ 4\sin }^{ 4 }x+{ { (1-{ \sin }^{ 2 }x) }^{ 2 } }=1$

${ 5\sin }^{ 4 }x={ 2\sin }^{ 2 }x$

Either,

${ 2\sin }^{ 2 }x=0,\\ \sin x=0,\\ x=n\pi.$

or,

$\sin x=\sqrt{\cfrac{5}{2}}\Longrightarrow$ Not possible.

So,

$x=n\pi.$

Solve the following equation:
$\displaystyle \sin^4 \, x \, + \, \cos^4 \, x \, = \, \sin \, x \, \cos \, x.$

${ ({ \sin }^{ 2 }x+{ \cos }^{ 2 }x) }^{ 2 }-{ 2\sin }^{ 2 }x{ \cos }^{ 2 }x=\sin x\cos x$

Let

$\sin x\cos x=z.$

$1-{2z}^{2}=z$

${2z}^{2}+z-1=0$

${2z}^{2}+2z-z-1=0$

$(z+1)(2z-1)=0$

So,

$z=-1,\\ \sin x\cos x=-1, \\ \sin 2x=-2.\text{[Thus no value possible.]}$

$2z-1=0,\\ \sin 2x=1, \\ 2x=2n\pi+\cfrac{\pi}{2},\\ x=n\pi+\cfrac{\pi}{4}.$

Solve the following equation:
$\displaystyle 2 \, \cos^2 \, x \, + \, 4 \, \cos \, x \, = \, 3 \,\sin^2 \, x$

${2\cos }^{2}x+4\cos x={3 \sin }^{2}x$

${2\cos }^{2}x+4\cos x=3(1-{\cos}^{2})$

$\cos x=\cfrac { -4\pm \sqrt { { 4 }^{ 2 }-4\times 5\times (-3) } }{ 2\times 5 }$

$\cos x=\cfrac { -4\pm \sqrt {16+60 } }{ 10 }=\cfrac { -4\pm \sqrt {76 } }{ 10 }$

So, let

$\cos x=\cfrac { -4\pm \sqrt {76 } }{ 10 }=p$

$\therefore$

$x=2n\pi\pm{\cos}^{-1}(p)$

Solve the following equations.
$\displaystyle (cos \, 6x \, - \, 1) \, cot \, 3x \, = \, sin \, 3x.$

$(cos 6x-1) cot 3x = sin3x$

$\Rightarrow (cos 6x- 1) \dfrac{cos3x}{sin3x} = sin3x$

$\Rightarrow (cos 6x-1) cos3x = sin^{2 }3x$

$\Rightarrow (2cos^2 3x-2) cos 3x = 1- cos^2 3x$

Substitute

$cos 3x = t$

$\Rightarrow 2(t^2 -1) t = 1-t ^2$

$\Rightarrow 2t^3 - 2t = 1-t^2$

$\Rightarrow 2t^3 + t^2 - 2t -1 = 0$

$t = 1$ is solution of the above equation

$\Rightarrow (t-1)(2t^2 +3t +1 ) = 0$

$\Rightarrow (t-1) (2t^2 +t+2t +1) = 0$

$\Rightarrow (t-1) [t (2t+1)+1(2t-1)] = 0$

$\Rightarrow (t-1) (t+1) (2t+1) = 0$

$\therefore t = 1$ or

$t = -1$ or

$t = -1/2$

$\Rightarrow cos3x =1$ or

$cos 3x = -1$ or

$cos 3x = -1/2$

$\Rightarrow 3x = 2n\pi \Rightarrow 3x = (2k+1)\pi \Rightarrow 3x = 2n\pi \pm \dfrac{2 \pi }{3}$

$x = \dfrac{2n\pi}{3} \Rightarrow x = (2k+1) \dfrac{\pi}{3} \Rightarrow x = \dfrac{2m\pi}{3}\pm \dfrac{2\pi^3}{9}$

$x \,\epsilon\, z$

$k\,\epsilon\, z$

$m\, \epsilon\, z$

1137936_887948_ans_e860d899c1414e779a992f4c1b7a6b8c.jpg

Solve: $\displaystyle \sin \, x \, + \, \sin \, 2x \, = \, \cos \, x \, + \, 2 \, \cos^2 \, x.$

$sinx+2sinxcosx=cosx+2cos^2x$

$sinx(1+2cosx)=cosx(1+2cosx)$

$\implies x=\pi/4,\pi/3$

Solve the following equation:
$\displaystyle 2 \, \cos^2 \, x \, - \, 1 \, = \, \sin \, 3x.$

$2(1-{ \sin }^{ 2 }x)-1=3\sin x-{ 4\sin }^{ 3 }x$

$2-{ 2\sin }^{ 2 }x-1-3\sin x+{ 4\sin }^{ 3 }x=0$

Let,

$\sin x=z.$

${ 4z }^{ 3 }-{ 2z }^{ 2 }-3z+1=0$

$(z-1)({ 4z }^{ 2 }+2z-1)=0$

So,

$z=1,\\ \sin x=1,\\ x=2n\pi+\cfrac{\pi}{2}.$

and

${ 4z }^{ 2 }+2z-1=0,\\ z=\cfrac { -2\pm \sqrt { 4+16 } }{ 8 }=\cfrac { -1\pm \sqrt { 5 } }{ 4 }.$

Let,

$a=\cfrac { -1+ \sqrt { 5 } }{ 4 }, \\b=\cfrac { -1- \sqrt { 5 } }{ 4 }.$

So,

$x=n\pi\pm{(-1)}^{n}{\sin}^{-1}(a)$ and

$x=n\pi\pm{(-1)}^{n}{\sin}^{-1}(b).$

Solve the following equations.
$\displaystyle 4 \, cos \, x \, - \, 2 \, cos \, 2x \, - \, cos \, 4x \, = \, 1.$

$4cos x - 2cos 2x - cos 4x =1$

$\Rightarrow 4 cos x - cos 2x - cos 4x = 1+cos 2x$

$\Rightarrow 4cosx -2 cos 3x cosx = 2cos^2x$

Say

$cos x = t$

$\Rightarrow 4t - 2t (4t^3-3t) = 2t^2$

$\Rightarrow 4t - 8t^4 + 6t^ 2 = 2t^2$

$\Rightarrow 8t^4 - 4t^2 - 4t = 0$

$\Rightarrow 2t^4 - t^2 -t = 0$

$\Rightarrow t (t-1) (2t^2 + 2t +1) = 0$

$\therefore t = 0$ or

$t = 1$

$\Rightarrow cosx = 0$ or

$\Rightarrow cosx = 1$

$\Rightarrow x = ( 2x+1)\pi/2 \Rightarrow x = 2 ,m\pi$

$n\epsilon$ integer

$m\epsilon$ integer

1137916_887923_ans_e9d32ab72eff4542a350362dc3413986.jpg

Solve the following equations.
$\displaystyle 8 \, cos \, x \, cos \, 2x \, cos \, 4x \, = \, \frac{sin \, 6x}{sin \, x}.$

$8cos x \, cos 2x\, cos 4x = \dfrac{sin 6x}{sinx}$

$\Rightarrow 8sinx. cosx. cos2x. cos\,4x = sin 6x$

$\Rightarrow 2sinx . cosx. 4cos2x . cos4x = sin6x$

$\Rightarrow 4 sin 2x \, cos 2x \, cos\, 4x = sin 6x$

$\Rightarrow 2(2sin2x. cos2x) cos 4x = sin6x$

$\Rightarrow 2 sin 4x. cos 4x = sin6x$

$\Rightarrow sin8x - sin6x = 0$

$\Rightarrow 2cos\, 7x. sinx = 0$

$\therefore cos 7x = 0 \,\,\, sinx = 0$

$7x = ( 2x+1) \pi/ 2 \,\,\, x = m\pi$

$x = (2x+1) \pi/4 \,\,\, m\epsilon z$

$x\epsilon z$

1137922_887934_ans_d96ca9646a094fe286254caa6966226d.jpg

Solve the following equation:
$\displaystyle \cos \, 4x \, + \, 2 \, \cos^2 \, x \, = \,1.$

${ 2\cos }^{ 2 }2x-1=1-{ 2\cos }^{ 2 }x=-\cos 2x$

Let

$\cos 2x=z.$

${2z}^{2}+z-1=0$

$(2z-1)(z+1)=0$

So,

$\cos 2x=-1$ and

$\cos 2x=\cfrac{1}{2}.$

$\therefore$

$x=(2n+1)\cfrac{\pi}{2}$ and

$x=n\pi\pm\cfrac{\pi}{6}.$

Solve the following equation:
$\displaystyle \sin^6 \, x \, + \, \cos^6 \, x \, = \, \frac{7}{16}.$

${ \sin }^{ 6 }x+{ (1-{ \sin }^{ 2 }x) }^{ 3 }=\cfrac { 7 }{ 16 }$

Let

$\sin x=z.$

${ z }^{ 6 }+1-{ z }^{ 6 }-{ 3z }^{ 2 }(1-{ z }^{ 2 })=\cfrac { 7 }{ 16 }$

$16-16({ 3z }^{ 2 }+{ 3z }^{ 4 })=7$

${48z}^{4}+{48z}^{2}-9=9$

${z}^{2}=\cfrac { -48\pm \sqrt { 4032 } }{ 96 } =\cfrac { -6\pm \sqrt { 63 } }{ 12 } .$

So,

${z}^{2}=\cfrac { -6\pm \sqrt { 63 } }{ 12 }=a$

$(let.)$

$\therefore$ So,

$x=n\pi\pm{(-1)}^{n}{\sin}^{-1}(\sqrt{a}).$

Solve the following equation:
$\displaystyle (1 \, - \, \tan \, x) \, (1 \, + \, \sin \, 2x) \, = \, 1 \, + \, \tan \, x$

$(1-\tan x)(1+\sin 2{x})=1+\tan x$

$1+\sin2{x}=\dfrac{1+\tan x}{1-\tan x}$

$(\sin x+\cos x)^{2}=\dfrac{\sin x+\cos x}{\cos x-\sin x}$

$\sin x+\cos x=0$ or

$\cos^{2}x-\sin^{2}x=1$

$\tan x=-1$ or

$\cos 2{x}=0$

$x=n\pi+\dfrac{3\pi}{4}$ or

$x=\dfrac{(2{m}+1)\pi}{4}$

Solve the following equation:
$\displaystyle \sin \, x \, + \, \sin^2 \, x \, + \, \cos^2 \, x \, = \, 0$

$\sin x=-1$

$x=2n\pi-\cfrac{\pi}{2}.$

Solve the following equations.
$\displaystyle sin \, \frac{3x}{2} \, cos \, \frac{x}{2} \, = \, \frac{sin \, 2x}{2}.$

$\sin { A } \cos { B } =\dfrac { 1 }{ 2 } \left( \sin { \left( A+B \right) } -\sin { \left( A-B \right) } \right)$

$\Rightarrow \sin { \dfrac { 3x }{ 2 } } \cos { \dfrac { x }{ 2 } } =\dfrac { 1 }{ 2 } \left[ \sin { 2x } -\sin { x } \right]$

$\Rightarrow \dfrac { \sin { 2x } }{ 2 } -\dfrac { \sin { x } }{ 2 } =\dfrac { \sin { 2x } }{ 2 }$ (given )

$\Rightarrow \dfrac { \sin { \left( -x \right) } }{ 2 } =0$

$\Rightarrow \sin { \left( -x \right) } =2n\pi$

$x=-2n\pi$

$\dfrac{{1 + \sin 2\theta + \cos 2\theta }}{{1 + \sin 2\theta - \cos 2\theta }} =$

$\rightarrow \dfrac { 1+\sin2\theta +\cos2\theta }{ 1+\sin2\theta -\cos2\theta }$

$\rightarrow \dfrac { 2\sin\theta \cos\theta +2{ \cos }^{ 2 }\theta }{ 2\sin\theta \cos\theta +2{ \sin }^{ 2 }\theta }$

$\rightarrow \dfrac { 2\cos\theta \left( \cos\theta +\sin\theta \right) }{ 2\sin\theta \left( \sin\theta +\cos\theta \right) }$

$\rightarrow \cot\theta$

$\dfrac{\tan A}{(1-\cot A)}+\dfrac{\cot A}{(1-\tan A)}=\sec A \cdot \text{cosec} A+1$

To Prove:

$\dfrac{\tan A}{(1-\cot A)}+\dfrac{\cot A}{(1-\tan A)}=\sec A \cdot cosec A+1$

Proof:

LHS

$\Rightarrow \dfrac{\tan A}{(1-\cot A)}+\dfrac{\cot A}{(1-\tan A)}$

Putiing

$\cot{A}=\dfrac{1}{\tan{A}}$

$\Rightarrow \dfrac{\tan A}{(1-\dfrac{1}{\tan{A}})}+\dfrac{\dfrac{1}{\tan{A}}}{(1-\tan A)}$

$\Rightarrow \dfrac{\tan^2{A}}{(\tan{A}-1)}+\dfrac{\dfrac{1}{\tan{A}}}{(1-\tan{A})}$

$\Rightarrow \dfrac{\tan^2{A}-\dfrac{1}{\tan{A}}}{\tan{A}-1}$

$\Rightarrow \dfrac{\tan^3{A}-1}{\tan{A}(\tan{A}-1)}$

$\Rightarrow \dfrac{(\tan{A}-1)^3+3\tan{A}(\tan{A}-1)}{\tan{A}(\tan{A}-1)}$

From

$(a-b)^3=a^3-b^3-3ab(a-b)\rightarrow a^3-b^3=(a-b)^3+3ab(a-b)$

Now,

$\dfrac{(\tan{A}-1)^2+3\tan{A}}{\tan{A}}$

$\dfrac{\tan^2{A}+1-2\tan{A}+3\tan{A}}{\tan{A}}$

$\dfrac{\tan^2{A}+1}{\tan{A}}+\dfrac{\tan{A}}{\tan{A}}$

$\dfrac{\sec^2{A}}{\sin{A}\sec{A}}+1$

$\sec{A} \cdot cosec{A}+1$ =RHS

If $\sin \theta-\sqrt{6} \cos \theta=\sqrt{7} \cos \theta$ . Prove that $\cos \theta-\sqrt{6}\sin \theta-\sqrt{7} \sin \theta=1$ .

$\rightarrow \sin\theta -\sqrt { 6 } \cos\theta =\sqrt { 7 } \cos\theta$

$\rightarrow \tan\theta =\sqrt { 6 } +\sqrt { 7 }$

$\rightarrow \cot\theta \left( \sqrt { 6 } +\sqrt { 7 } \right) =1$

$\rightarrow \sqrt { 6 } \sin\theta +\sqrt { 7 } \sin\theta =\cos\theta$

$\rightarrow \cos\theta -\sqrt { 6 } \sin\theta -\sqrt { 7 } \sin\theta =1$

If $\cot \theta=\sqrt {7}$ , find the value of $\dfrac {cosec^{2} \theta - \sec^{2}\theta}{cosec^{2} \theta + \sec^{2}\theta}$ . $[\because \cot \theta=\dfrac{1}{ \tan \theta}]$

$\tan\theta =\dfrac { 1 }{ \sqrt { 7 } }$

$\dfrac { \text{ cosec }^{ 2 }\theta -{ \sec }^{ 2 }\theta }{ \text{ cosec }^{ 2 }\theta +{ \sec }^{ 2 }\theta }$

$= \dfrac { 1-{ \tan }^{ 2 }\theta }{ 1+{ \tan }^{ 2 }\theta }$

$= \dfrac { 1-\dfrac { 1 }{ 7 } }{ 1+\dfrac { 1 }{ 7 } }$

$=\dfrac { 6 }{ 8 }$

$=\dfrac { 3 }{ 4 }$

Prove that : $\tan 2x=\dfrac{2\tan x}{1-\tan^2x}$

$\tan { (2x) } \quad =\quad \cfrac { 2\tan { (x) } }{ 1-\tan ^{ 2 }{ x } } \\ LHS\quad =\quad tan\{ (2x)\\ =\quad \tan { (x+x) } \\ =\quad \cfrac { \tan { (x)+\tan { (x) } } }{ 1-\tan { (x) } \tan { (x) } } \\ =\cfrac { 2\tan { (x) } }{ 1-\tan ^{ 2 }{ x } } \quad =\quad RHS$

If $A+B+C = \pi$ then prove that, $\cos 2A+\cos 2B-\cos 2C=1-4\sin A\sin B\cos C$ .

Here we want

$+1$ and as such we write

$\cos 2A=1-2\sin^2 A$ and combine the other two terms.
L.H.S.

$=1-2\sin^2A+2\sin(B+C)\sin (C-B)$ (Note)

$=1-2\sin^2A-2\sin A\sin (B-C)$

$[\because \sin (-\theta)=-\sin\theta]$

$=1-2\sin A[\sin A+\sin (B-C)]$

$=1-2\sin A[\sin (B+C)+\sin(B-C)]$

$=1-2\sin A(2\sin B\cos C)$

$=1-4\sin A\sin B\cos C$ .

$\cos \dfrac{\pi}{15}\cos \dfrac{2\pi}{15}\cos \dfrac{3\pi}{15}\cos \dfrac{4\pi}{15}\cos \dfrac{5\pi}{15}\cos \dfrac{6\pi}{15}\cos \dfrac{7\pi}{15}=\dfrac{1}{2^7}$

1312057_1004686_ans_cd57d1b4e31b47ab90fc5038547644bc.jpg

Show that:
$\cos^248^{\circ}-\sin^212^{\circ}=\dfrac{\sqrt{5}+1}{8}$

$\cos ^{ 2 }{ 48-\sin ^{ 2 }{ 12 } } \\ =\quad \cos { (48+12) } \cos { (48-12) } \\ =\quad \cos { (60^0) } \cos { (36^0) } \\ =\quad \cfrac { 1 }{ 2 } \times \cfrac { \sqrt { 5 } +1 }{ 4 } \\ =\quad \cfrac { \sqrt { 5 } +1 }{ 8 }$

The value of $\cos^2 \, A \, + \, \cos^2 \, B \, - \, 2 \, \cos \, A \, \cos \, B \, \cos \, (A \, + \, B) \, -\, \sin^2 \, (A \, + \, B)$

L.H.S. =

$cos^2 \, A \, + \, cos^2 \, B \, - \, 2 \, cos \, A \, cos \, B \, (cos \, A \, cos \, B \, - \, sin \, A sin \, B)$

$= \, (cos^2 \, A \, - \, cos^2 \, A \, cos^2 \, B) \, + \, (cos^2 \, B \, - \, cos^2 \, A \, cos^2 \, B) \, + \, 2 \, sin \, A \, sin \, B \, cos \, A \, cos \, B$

$= \, cos^2 \, A \, (1 \, - \, cos^2 \, B) \, + \, cos^2 \, B \, (1 \, - \, cos^2 \, A) \, + \, 2 \, sin \, A \, sin \, B \, cos \, A \, cos \, B$

$= \, cos^2 \, A \, sin^2 \, B \, + \, cos^2 \, B \, sin^2 \, A \, + \, 2 \, sin \, A \, sin \, B \, cos \, A \, cos \, B$

$= \, (sin \, A \, cos \, B \, + \, cos \, A \, sin \, B)^2 \, = \, sin^2 \, (A \, + \, B)$

Find the real roots of the equation $\cos ^{ 7 }{ x } +\sin ^{ 4 }{ x } =1$ in the interval $\left( -\pi ,\pi \right)$ .

Ans.

$-\pi/2,0,\pi/2$

$cos^{ 7 }x=1-\sin^{ 4 }x=(1-\sin^{ 2 }x)(1+\sin^{ 2 }x)$

$=\cos^{ 2 }x(1+\sin^{ 2 }x)$

$\therefore\quad \cos x=0$ or

$x=\pi/2,-\pi/2$
or

$\cos^{ 5 }x=1+\sin^{ 2 }x$ or

$\cos^{ 5 }x-\sin^{ 2 }x=1$
Now maximum value of each

$\cos x$ or

$\sin x$ is

$1$ . Hence the above equation will hold when

$\cos x=1$ and

$\sin x=0$ . Both these imply

$x=0$ .
Hence

$x=\dfrac{ \pi }{ 2 },\dfrac{ \pi }{ 2 },0$

Prove that $\sin^2(\dfrac{A}{2})+\sin^2(\dfrac {B}{2})+\sin^2(\dfrac {C}{2})=1-2\sin(\dfrac {A}{2})\sin\dfrac {B}{2})\sin(\dfrac {C}{2})$ .

We want

$+1$ in R.H.S.

$\therefore$ L.H.S.

$=1-\cos^2(A/2)+\sin^2(B/2)+\sin^2(C/2)$

$=1-\{\cos^2(A/2)-\sin^2(B/2)\}+\sin^2(C/2)$

$=1-\cos\{(A+B)/2\}\cos \{(A-B)/2\}+\sin^2(C/2)$

$=1-\sin(C/2)\cos \{(A-B)/2\}+\sin^2(C/2)$

$=1-\sin(C/2)[\cos \{(A-B)/2\}-\sin (C/2)]$

$=1-\sin(C/2)[\cos \{(A-B)/2\}-\cos \{(A+B)/2\}]$

$=1-\sin (C/2)[2\sin (A/2)\sin (B/2)]$

$=1-2\sin(A/2)\sin (B/2)\sin (C/2)$ .

In any triangle ABC prove the identities.
$\sin^2A+\sin^2B-\sin^2C=2\sin A\sin B\cos C$ .

L.H.S.

$=\sin^2A+\sin(B+C)\sin(B-C)$

$=s\in^2A+\sin A\sin(B-C)$

$=\sin A[\sin A+\sin (B-C)]$

$=\sin A[\sin (B+C)+\sin (B-C)]$

$=\sin A[2\sin B\cos C]$

$=2\sin A\sin B\cos C$ .

If A+B+C $=\pi$ Prove that: $\dfrac{\sin 2A+\sin 2B+\sin 2C}{\sin A+\sin B+\sin C}=8\sin \bigg(\dfrac{A}{2}\bigg)\sin \bigg(\dfrac{B}{2}\bigg)\sin \bigg(\dfrac{C}{2}\bigg)$ .

L.H.S.

$=\dfrac{4\sin A\sin B\sin C}{4\cos (A/2)\cos (B/2)\cos (C/2)}$

$=\dfrac{\{2\sin (A/2)\cos (A/2)\}\{2\sin (B/2)\cos (B/2)\}\{(2\sin (C/2)\cos (C/2)\}}{\cos (A/2)\cos (B/2)\cos (C/2)}$

$= 8\sin (\dfrac{A}{2})\sin (\dfrac{B}{2})\sin (\dfrac{C}{2})$ .

In any triangle ABC prove that the identities.
$\dfrac{\sin 2A+\sin 2B+\sin 2C}{\cos A+\cos B+\cos C-1}=8\cos(A/2)\cos (B/2)\cos (C/2)$ .

L.H.S.

$=\dfrac{4\sin A\sin B\sin C}{4\sin (A/2)\sin (B/2)\sin (C/2)}$

$=8\cos(A/2)\cos(B/2)\cos(C/2)$

$\because \sin A=2\sin(A/2)\cos (A/2)$ etc.

If $A+B+C=2k$ , then prove that $\cos^2k+\cos^2(k-A)+\cos^2(k-B)+\cos^2(k-C)=2+2\cos A\cos B\cos C$ .

$\cos^2x-\sin^2y=\cos(x+y)\cos (x-y)$
L.H.S.

$=2+[\cos^2k-\sin^2(k-A)]+[\cos^2(k-B)-\sin^2(k-C)]$

$=2+\cos (2k-A)\cos A+\cos (2k-B-C)\cos (C-B)$
Put

$2k-A=B+C$ and

$2k-B-C=A$

$=2+\cos A[\cos (B+C)+\cos (C-B)]$

$=2+2\cos A\cos B\cos C$ .

Prove that: $\sin^2A+\sin^2B+\sin^2C=2+2\cos A\cos B\cos C$ .

On changing

$\sin^2A$ to

$1-\cos^2A$ etc.

Proceeding directly as we need

$2$ , we write

$\sin^2A=1-\cos^2A$ and

$\sin^2B=1-\cos^2B$

$L.H.S. =2-\cos^2A-\cos^2B+\sin^2C$

$=2-(\cos^2A-\sin^2C)-cos^2B$

$=2-\cos(A+C)\cos (A-C)-\cos^2B$

$=2+\cos B\cos (A-C)-\cos^2B$

$=2+\cos B[\cos (A-C)-\cos B]$

$=2+\cos B[\cos (A-C)+\cos (A+C)]$

$=2+\cos B(2\cos A\cos C)$

$=2+2\cos A\cos B\cos C$ .

Express $\sin 3A+\sin 3B+\sin 3C$ as the product of three trigonometrical ratios where A, B, C are the angles of a triangle. If the given expression be zero, then at least one angle of the triangle is $60^o$

$A+B+C=\pi, \therefore 3A+3B+3C=3\pi$

$\dfrac{1}{2}(3A+3B)=\dfrac{3}{2}\pi -\dfrac{3}{2}C$

$\therefore \sin\dfrac{1}{2}(3A+3B)=-\cos \dfrac{3}{2}C$
and

$\cos \dfrac{1}{2}(3A+3B)=-\sin\dfrac{3}{2}C$ .
L.H.S.

$=2\sin\dfrac{3A+3B}{2}\cos \dfrac{3A-3B}{2}+2\sin \dfrac{3C}{2}\cos \dfrac{3C}{2}$

$=-2\cos \dfrac{3C}{2}\cos \dfrac{3A-3B}{2}+2\sin \dfrac{3C}{2}\cos \dfrac{3C}{2}$

$=-2\cos \dfrac{3C}{2}\left[\cos \dfrac{3A-3B}{2}-\sin \dfrac{3C}{2}\right]$

$=-2\cos \dfrac{3C}{2}\left[\cos \dfrac{3A-3B}{2}+\cos \dfrac{3A+3B}{2}\right]$

$=-4\cos \dfrac{3A}{2}\cos \dfrac{3B}{2}\cos \dfrac{3C}{2}$
If it be zero then at least one factor is zero, hence

$3A/2=90^o$ or

$A=60^o$ etc.

If $A+B+C=\pi$ , prove that $(\sin A+\sin B+\sin C)(-\sin A+\sin B+\sin C)(\sin A-\sin B+\sin C)(\sin A+\sin B-\sin C) \\ =4\sin^2A\sin^2B\sin^2C$ .

Then L.H.S.

$=\left[4\cos\dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2}\right]\left[4\sin \dfrac{B}{2}\sin \dfrac{C}{2}\cos \dfrac{A}{2}\right]$

$\cdot \left[4\sin\dfrac{C}{2}\sin \dfrac{A}{2}\cos \dfrac{B}{2}\right]\left[4\sin\dfrac{A}{2}\sin \dfrac{B}{2}\cos \dfrac{C}{2}\right]$

$=4\times 64\left(\sin^2\dfrac{A}{2}\cos^2\dfrac{A}{2}\right)$

$( ) ( )$

$=4\left(2\sin \dfrac{A}{2}\cos \dfrac{A}{2}\right)^2$

$( ) ( )$

$=4\sin^2A\sin^2B\sin^2C$ .

Prove that: $\sin^2(\dfrac {A}{2})+\sin^2(\dfrac {B}{2})-\sin^2(\dfrac {C}{2})=1-2\cos(\dfrac {A}{2})\cos(\dfrac {B}{2})\sin(\dfrac{C}{2})$ .

We want

$+1$ in R.H.S.
L.H.S.

$=1-\cos^2(A/2)+\sin^2(B/2)-\sin^2(C/2)$

$=1-\cos^2(A/2)+[\sin \{(B+C)/2\}\sin \{(B-C)/2\}]$

$=1-\cos (A/2)[\cos A/2-\{\sin (B-C)/2\}]$

$=1-\cos (A/2)[\sin \{(B+C)/2-\sin (B-C)/2\}]$

$=1-\cos (A/2)[2\cos (B/2)\sin (C/2)]$

$=1-2\cos (A/2)\cos (B/2)\sin (C/2)$ .

Prove that: $\cos^2A+\cos^2B+\cos^2C=1-2\cos A\cos B\cos C$ .

We write

$\cos^2A=1-\sin^2A$

and as in

$\Delta ABC$

$A+B+C=180$

$\implies \cos C=\cos (180-A-B)=-\cos (A+B)$

$L.H.S.=1-\sin^2A+\cos^2B+\cos^2C$

$=1+(\cos^2B-\sin^2A)+\cos^2C$

$=1+\cos(B+A)\cos (B-A)+\cos^2 C$ .....

$(\cos^2C-\sin^2D=\cos(C+D)\cos (C-D))$

$=1-\cos C\cos (B-A)+\cos^2C$

$=1-\cos C[\cos (B-A)-\cos C]$

$=1-\cos C[\cos (B-A)+\cos (B+A)]$ .

$=1-\cos C(2\cos A\cos B)$ ....... By compound angles formula

$=1-2\cos A\cos B\cos C$ .

Solve $2\sin^2x+\sqrt{3}\cos x+1=0$ .

Changing

$\sin^2x$ into

$1-\cos^2x$ ; we get

$2(1-\cos^2x)+\sqrt{3}\cos x+1=0$
or

$2\cos^2x-\sqrt{3}\cos x-3=0$

$\therefore \cos x=\dfrac{\sqrt{3}\pm \sqrt{3+24}}{4}$

$=\dfrac{\sqrt{3}+3\sqrt{3}}{4}=\sqrt{3}$ or

$\dfrac{-\sqrt{3}}{2}$
Since

$\sqrt{3}$ is greater than

$1$ it is not admissible as

$\cos x$ can not be greater than

$1$ .

$\therefore \cos x=-\sqrt{3}/2=-\cos (\pi /6)$

$=\cos (\pi -\pi/6)=\cos (5\pi /6)$

$\therefore x=2n\pi \pm (5\pi /6)$ .

Find m if $\sin { A } \sin { \left( { 60 }^{ o }-A \right) } \sin { \left( { 60 }^{ o }+A \right) } =\cfrac { 1 }{ m} \sin { 3A } \quad$

$LHS=\sin { A } \sin { (60-A) } \sin { (60+A) } \\ =\sin { A } [\sin { ^{ 2 }60 } -\sin { ^{ 2 }A } ]\\ =\sin { A } [\cfrac { 3 }{ 4 } -\sin { ^{ 2 }A } ]\\ =\cfrac { \sin { A } }{ 4 } [3-4\sin { ^{ 2 }A } ]\\ =\cfrac { 3\sin { A } -4\sin { ^{ 3 }A } }{ 4 } \\ =\cfrac { \sin { 3A } }{ 4 } \\ =RHS$

If $x+y+z=xyz$ , prove that $\dfrac{x+y}{1-xy}+\dfrac{y+z}{1-yz}+\dfrac{z+x}{1-zx}=\dfrac{x+y}{1-xy}\cdot \dfrac{y+z}{1-yz}\cdot \dfrac{z+x}{1-zx}$ .

As above

$A+B+C=0$ or

$n\pi$

$\tan A+\tan B+\tan C=\tan A\tan B\tan C$ ..

$(1)$
Now

$A=\pi -(B+C)$

$\therefore \tan A=-\tan(B+C)$
or

$\tan A=-\dfrac{\tan B+\tan C}{1-\tan B\tan C}=-\left(\dfrac{y+z}{1-yz}\right)$ etc.
Now put in

$(1)$ and cancel

$-$ ive sign from both sides.

$\therefore \dfrac{x+y}{1-xy}+\dfrac{y+z}{1-yz}+\dfrac{z+x}{1-zx}=\dfrac{x+y}{1-xy}\cdot \dfrac{y+z}{1-yz}\cdot \dfrac{z+x}{1-zx}$ .

If $x+y+z=xyz$ , then prove that.
For any three angles A, B, C, $\tan(B-C)+\tan(C-A)+\tan(A-B)=\tan(B-C)\tan(C-A)\tan(A-B)$ .

Put

$B-C=p, C-A=q, A-B=r$

$\therefore p+q+r=0$
or

$\tan (p+q+r)=\tan 0^o=0$
or

$\dfrac{S_1-S_3}{1-S_2}=0$

$\therefore S_1=S_3$
or

$\tan p+\tan q+\tan r=\tan p\tan q\tan r$
Now put the values of p, q and r.

$\therefore \tan(B-C)+\tan(C-A)+\tan(A-B)=\tan(B-C)\tan(C-A)\tan(A-B)$ .

Solve: $5\cos 2\theta +2\cos^2\left(\dfrac{1}{2}\theta \right)+1=0, -\pi < \theta < \pi$ .

Given,

$5(2\cos^2\theta -1)+(1+\cos\theta)+1=0$

$10\cos^2\theta +\cos \theta -3=0$

$10\cos^2\theta +6\cos \theta-5\cos \theta -3=0$

$(5\cos \theta +3)(2\cos \theta -1)=0$

$\therefore \theta =\pi/3, -\pi/3, \pi -\cos^{-1}(3/5)$

$\because -\pi < \theta < \pi$ .

Solve $4 \cos \theta - 3 \sec \theta = 2 \tan \theta$ .

$4\cos \theta -\dfrac{3}{\cos \theta }=\dfrac{2\sin \theta}{\cos \theta}$

$4(1-\sin^2\theta)=2\sin\theta +3$

$4\sin^2\theta +2\sin\theta -1=0$

$\therefore \sin\theta =\dfrac{-2\pm \sqrt{(4+16)}}{8}$

$=\dfrac{-2\pm 2\sqrt{5}}{8}$

$=\dfrac{\sqrt{5}-1}{4}, -\left(\dfrac{\sqrt{5}+1}{4}\right)$

$\sin\theta =\sin 18^o$ i.e,

$\sin(\pi /10)$ ,

$\therefore \theta =n\pi +(-1)^n(\pi/10)$

$\sin\theta =-\cos 36^o=-\sin 54^o$

$=\sin (-3\pi/10)$

$\therefore \theta =n\pi -(-1)^n(3\pi /10)$ .

If $x+y+z=xyz$ , prove that $\dfrac{3x-x^3}{1-3x^2}+\dfrac{3y-y^3}{1-3y^2}+\dfrac{3z-z^3}{1-3z^2}=\dfrac{3x-x^3}{1-3x^2}\cdot \dfrac{3y-y^3}{1-3y^2}\cdot \dfrac{3z-z^3}{1-3z^2}$ .

$\tan (3A+3B+3C)=0$

$\therefore \dfrac{S_1-S_3}{1-S_2}=0$ so that

$S_1=S_3$
or

$\tan 3A+\tan 3B+\tan 3C=\tan 3A\cdot \tan 3B\cdot 3C$
Now put

$\tan 3A=\dfrac{3\tan A-\tan^3A}{1-3\tan^2A}=\dfrac{3x-x^3}{1-3x^2}$ etc.

SImilarly we can write for

$\tan 3B$ and

$\tan 3C$ .

$\therefore \dfrac{3x-x^3}{1-3x^2}+\dfrac{3y-y^3}{1-3y^2}+\dfrac{3z-z^3}{1-3z^2}=\dfrac{3x-x^3}{1-3x^2}\cdot \dfrac{3y-y^3}{1-3y^2}\cdot \dfrac{3z-z^3}{1-3z^2}$ .

The number of values of $\theta$ satisfying $\sin \theta +\sin 3\theta +\sin 5\theta =0; 0\leq \theta \leq \dfrac{1}{2}\pi$ is

1909461_1006224_ans_fdb88a9127a5487887281a495d502526.jpg

Solve $r\sin \theta =\sqrt{3}, r+4\sin \theta =2(\sqrt{3}+1), 0\leq \theta \leq 2\pi$ .

Eliminating r between the given equations, we get

$\dfrac{\sqrt{3}}{\sin\theta}+4\sin\theta =2(\sqrt{3}+1)$
or

$4\sin^2\theta -2\sqrt{3}\sin\theta -2\sin\theta +\sqrt{3}=0$

$2\sin\theta (2\sin \theta -\sqrt{3})-1.(2\sin\theta -\sqrt{3})=0$

$\therefore (2\sin\theta -1)(2\sin \theta -\sqrt{3})=0$

$\sin\theta =\dfrac{1}{2}=\sin\dfrac{\pi}{6}$ or

$\sin\left(\pi -\dfrac{\pi}{6}\right)$

$\sin\theta =\dfrac{\sqrt{3}}{2}=\sin\dfrac{\pi}{3}$ or

$\sin\left(\pi -\dfrac{\pi}{3}\right)$

$\therefore \theta =\pi/6, \pi/3, 5\pi/6, 2\pi/3, 0\leq \theta \leq 2\pi$ .

The smallest $+$ ve x satisfying the equation $log_{cos x}\sin x+\log_{\sin x}\cos x=2$ is $\pi /4$ .

$y+\dfrac{1}{y}=2$

$\Rightarrow (y-1)=0$

$\Rightarrow y=1$

$\therefore log_{\cos x}\sin x=1$ or

$\sin x=\cos x$
or

$\tan x=1$

$\therefore x=\pi/4$ .

Solve $\sin \theta +\sin 2\theta +\sin 3\theta +\sin 4\theta =0$ .

$(\sin \theta +\sin 3\theta)+(\sin 2\theta +\sin 4\theta)=0$

$2\sin 2\theta \cos \theta +2\sin 3\theta \cos \theta =0$

$2\cos \theta (\sin 2\theta +\sin 3\theta)=0$

$4\cos \theta \sin (5\theta/2)\cos (\theta/2)=0$

$\cos\theta =0=\cos (\pi/2)$ ,

$\therefore \theta =\left(n+\dfrac{1}{2}\right)\pi$

$\cos (\theta/2)=0=\cos (\pi/2)$ ,

$\therefore \dfrac{\theta}{2}=\left(n+\dfrac{1}{2}\right)\pi$ or

$\theta =(2n+1)\pi$

$\sin (5\theta /2)=0=\sin 0$

$\therefore (5\theta /2)=n\pi$ or

$\theta =2n\pi/5$ .

Solve $3\tan^2\theta -2\sin \theta =0$ .

$3\cdot \dfrac{\sin^2\theta}{\cos^2\theta}-2\sin \theta =0, \cos\theta \neq 0$ .
Multiplying throughout by

$\cos^2\theta$ and then changing it into

$\sin^2\theta$ . we get

$3\sin^2\theta -2\sin\theta (1-\sin^2\theta)=0$

$\sin\theta (3\sin \theta -2+2\sin^2\theta)=0$

$\sin\theta (2\sin^2\theta +4\sin\theta -\sin\theta -2)=0$

$\sin\theta (\sin \theta +2)(2\sin\theta -1)=0$

$\therefore \sin\theta =0$ ,

$\therefore \theta =n\pi$

$\sin\theta =1/2=\sin(\pi/6)$ ,

$\therefore \theta =n\pi +(-1)^n\pi/6$

$\sin\theta =-2$ (rejected as

$\sin\theta$ cannot be greater than one numerically).

Solve $\cot \theta -\tan \theta =\sec \theta$ .

$\dfrac{\cos\theta}{\sin\theta}-\dfrac{\sin\theta}{\cos\theta}=\dfrac{1}{\cos\theta}$
or

$\dfrac{\cos 2\theta}{\sin\theta \cos\theta}=\dfrac{1}{\cos\theta}$

$\therefore 1-2\sin^2\theta =\sin\theta$

$(\cos\theta \neq 0)$
or

$2\sin^2\theta +\sin\theta -1=0$
or

$(2\sin\theta -1)(\sin\theta +1)=0$

$\therefore \sin\theta =\dfrac{1}{2}$

$\therefore \theta =n\pi+(-1)^n\dfrac{\pi}{6}$
But

$\sin\theta =-1$ implies

$\cos\theta =0$ but

$\cos\theta \neq 0$
Hence no value will be admissible when

$\sin\theta =-1$ .

Solve $\cos^{40}\theta +\sin^{58}\theta =1$ .

Since both

$\sin\theta$ and

$\cos\theta$ are less then unity, both terms are

$+$ ive the above result in possible only if either

$\sin\theta =1$ i.e.,

$\cos\theta =0$
or

$\cos \theta =1$ ,

$\sin\theta =0$

$\cos\theta =0, \theta =\left(n+\dfrac{1}{2}\right)\pi =(2n+1)\dfrac{\pi}{2}=$ odd

$\dfrac{\pi}{2}$ ..

$(1)$

$\sin \theta =0, \theta =n, \pi =2n\cdot \dfrac{\pi}{2}=$ even

$\dfrac{\pi}{2}$ .

$(2)$

$\therefore \theta =\dfrac{r\pi}{2}$ , where

$r\in I$ by

$(1)$ and

$(2)$ .

Solve $sec 4\theta -sec 2\theta =2$ .

$\dfrac{1}{\cos 4\theta}-\dfrac{1}{\cos 2\theta}=2$

$\cos 4\theta \neq 0, \cos 2\theta \neq 0$ .

$(1)$

$\therefore \cos 2\theta -\cos 4\theta =2\cos 2\theta \cos 4\theta$

$=\cos 2\theta +\cos 6\theta$

$\therefore \cos 6\theta +\cos 4\theta =0$
or

$2\cos 5\theta \cos \theta =0$

$\cos \theta =0=\cos (\pi/2)$

$\therefore \theta =n\pi +\pi/2$

$\cos 5\theta =0=\cos (\pi/2)$

$\therefore 5\theta =n\pi +\pi/2$
or

$\theta =\dfrac{n\pi}{5}+\dfrac{\pi}{10}$ .

Solve $\dfrac{\sqrt{3}}{2}\sin x-\cos x=cos^2x$ .

$\dfrac{1}{2}\sqrt{3}\sin x=\cos x+\cos^2x$ . Square

$3(1-\cos^2x)=4(\cos^2x+2\cos^3x+\cos^4x)$
or

$4\cos^4x+8cos^3+7\cos^2x-3=0$
Clearly

$\cos x=-1$ satisfies above, hence it can be written as

$(\cos x+1)(4\cos^3x+4\cos^2x+3\cos x-3)=0$
Again

$\cos x=\dfrac{1}{2}$ satisfies the

$2$ nd factor

$(\cos x+1)(2\cos x-1)(2\cos^2x+3\cos x+3)=0$

$\therefore \cos x=-1=\cos \pi$

$\therefore x=2n\pi +\pi =(2n+1)\pi$

$\cos x=1/2=\cos (\pi /3)$

$\therefore x=2n\pi \pm \pi/3$

$2\cos^2x+3\cos x+3=0$ ,

$B^2-4AC=9-4.2.3=-$ ive.
Hence this factor does not give any real values of

$\cos x$ .

Solve $\tan 3\theta +\tan \theta =2\tan 2\theta$ .

$\tan 3\theta +\tan \theta =2\tan 2\theta$

$\tan 3\theta -\tan 2\theta =\tan 2\theta -\tan \theta$

$\dfrac{\sin (3\theta -2\theta)}{\cos 3\theta \cos 2\theta}=\dfrac{\sin (2\theta -\theta)}{\cos 2\theta \cos \theta}$
or

$\sin \theta (\cos \theta -\cos 3\theta )=0$
or

$\sin \theta \cdot 2\sin \theta \sin 2\theta =0$

$\therefore \sin\theta =0$ ,

$\therefore \theta =n\pi$

$\sin 2\theta =0$ ,

$\therefore 2\theta =n\pi$ or

$\theta =\dfrac{n\pi}{2}$
But

$\theta =\dfrac{n\pi}{2}$ is rejected when n is odd as it makes

$\infty =\infty$ .

$\therefore \theta =n\pi$ as it will make

$0=0$ .

Prove that:
$\dfrac{1}{1 \, + \, \sin \, \theta} \, + \, \dfrac{1}{1 \, - \, \sin \, \theta} \, = \, 2 \, \sec^2 \, \theta$

We have,
LHS = $\dfrac{1}{1 \, + \, \sin \, \theta} \, + \dfrac{1}{1 \, - \, \sin \, \theta}$

$\Rightarrow$ LHS = $\dfrac{1 \, - \, \sin \, \theta \, + \, 1 \, + \, \sin \, \theta}{(1 \, + \, \sin \, \theta)(1 \, - \, \sin \, \theta)}$

$\Rightarrow$ LHS = $\dfrac{2}{1 \, - \, \sin ^2 \, \theta}$

$\Rightarrow$ LHS = $\dfrac{2}{\cos ^2 \, \theta}$ $[\because \, 1 \, - \, \sin ^2 \, \theta \, = \, \cos ^2 \, \theta]$

$\Rightarrow$ LHS = $2 \, sec^2 \, \theta$ = RHS $\left[\because \, sec \, \theta \, = \, \dfrac{1}{\cos \, \theta} \right]$

Prove that :
$\dfrac{\tan \, \theta \, + \, \sin \, \theta}{\tan \, \theta \, - \, \sin \, \theta} \, = \, \dfrac{\sec \, \theta \, + \, 1}{\sec \, \theta \, - \, 1}$

We have,

L.H.S. = $\dfrac{tan \, \theta \, + \, \sin \, \theta}{tan \, \theta \, - \, \sin \, \theta}$

$\Rightarrow \, \,$ L.H.S. = $\dfrac{\dfrac{\sin \, \theta}{\cos \, \theta} \, + \, \sin \, \theta}{\dfrac{\sin \, \theta}{\cos \, \theta} \, - \, \sin \, \theta}$

$\Rightarrow \, \,$ L.H.S. = $\dfrac{\sin \, \theta \left(\dfrac{1}{\cos \, \theta} \, + \, 1\right)}{\sin \, \theta \left(\dfrac{1}{\cos \, \theta} \, - \, 1\right)} \, = \, \dfrac{\dfrac{1}{\cos \, \theta} \, + \, 1}{\dfrac{1}{\cos \, \theta} \, - \, 1} \, = \, \dfrac{sec \, \theta \, + \, 1}{sec \, \theta \, - \, 1} \, = \, R.H.S.$

${\sin ^2}n\theta - {\sin ^2}(n - 1)\theta = {\sin ^2}\theta$ , where $n$ is constant and $n \ne 0,1$

$LHS=\sin { ^{ 2 }{ n\theta } } -\sin { ^{ 2 }(n-1)\theta } \\ =\sin { [n-(n-1)]\theta } .\sin { [n+(n-1)]\theta } \\ =\sin { \theta } .\sin { [2n-1]\theta } \\ \therefore \sin { ^{ 2 }\theta } =\sin { \theta } .\sin { [2n-1]\theta } \\ \therefore \sin { \theta } =\sin { [2n-1]\theta } \\ \therefore n=\cfrac { 1 }{ 2 }$

Solve the equation $\sin x+\cos x=\sin 2x-1$ .

$\sin x+\cos x=t$ then on squaring

$1+\sin 2x=t^2$

$\therefore t=(t^2-1)-1$ or

$t^2-t-2=0$
or

$(t-2)(t+1)=0$

$\therefore t=2, -1$

$\cos x+\sin x=2$ or

$-1$
or

$\dfrac{1}{\sqrt{2}}\cos x+\dfrac{1}{\sqrt{2}}\sin x=\sqrt{2}$ or

$-\dfrac{1}{\sqrt{2}}$
or

$\cos\left(x-\dfrac{\pi}{4}\right)=\sqrt{2}$ or

$-\cos\dfrac{\pi}{4}$
The first option is rejected as

$\sqrt{2} > 1$

$\therefore \cos\left(x-\dfrac{\pi}{4}\right)=\cos \left(\pi -\dfrac{\pi}{4}\right)=\cos \dfrac{3\pi}{4}$

$\therefore x-\dfrac{\pi}{4}=2n\pi\pm \dfrac{3\pi}{4}$

$\therefore x=2n\pi +\pi$ or

$2n\pi -\dfrac{\pi}{2}$ .

Prove that :
$\dfrac{tan \, \theta \, - \, cot \, \theta}{sin \, \theta \, cos \, \theta} \, = \, sec^2 \, \theta \, - \, cosec^2 \, \theta \, = \, tan^2 \, \theta \, - \, cot^2 \, \theta$

We have,
LHS =

$\dfrac{tan \, \theta \, - \, cot \, \theta}{sin \, \theta \, cos \, \theta} \, = \, \dfrac{\dfrac{sin \, \theta}{cos \, \theta} \, - \, \dfrac{cos \, \theta}{sin \, \theta}}{sin \, \theta \, cos \, \theta} \, = \, \dfrac{\dfrac{sin^2 \, \theta \, - \, cos^2 \, \theta}{sin \, \theta \, cos \, \theta}}{cos \, \theta \, sin \, \theta}$

$\Rightarrow$ LHS =

$\dfrac{sin^2 \, \theta \, - \, cos^2 \, \theta}{sin^2 \, \theta \, cos^2 \, \theta}$

$\Rightarrow$ LHS =

$\dfrac{sin^2 \, \theta}{sin^2 \, \theta \, cos^2 \, \theta} \, - \, \dfrac{cos^2 \, \theta}{sin^2 \, \theta \, cos^2 \, \theta}$

$\Rightarrow$ LHS =

$\dfrac{1}{cos^2 \, \theta} \, - \, \dfrac{1}{sin^2 \, \theta}$

$\Rightarrow$ LHS =

$sec^2 \, \theta \, - \, cosec^2 \, \theta \, = \, (1 \, + \, tan^2 \, \theta) \, - \, (1 \, + \, cot^2 \, \theta) \, = \, tan^2 \, \theta \, - \, cot^2 \, \theta \, = \, RHS.$

Prove that :
$\dfrac{(1 \, + \, cot \, A \, + \, tan \, A)(sin \, a \, - \, cos \, A)}{sec^3 \, A \, - \, cosec^3 \, A} \, = \, sin^2 \, A \, cos^2 \, A$

$LHS = \dfrac { \left ( 1 \, + \, \dfrac {cos A}{sin A} \, + \, \dfrac {sin A }{cos A} \right) (sin A \, - \, cosA )}{ \left ( \dfrac {1 }{cos^3 A} \, - \, \dfrac {1}{sin^3 A} \right)}$

$\Rightarrow \, \,. \, LHS \, = \, \dfrac { \left ( 1 \, + \, \dfrac {cos^2 A\, + \, sin ^2 \, A}{sin A \, cos A}\right) (sin A \, - \, cos A)}{\left ( \dfrac {sin^3 A \, - \, cos^3 \, A }{sin ^3 A \, cos^3 A}\right) }$

$\Rightarrow \, \, \, \,LHS \, = \, \dfrac {\left ( 1 \, + \, \dfrac {1}{sinA \, cos A} \right ) (sin A \, - \, cos A ) sin^3 A \, cos^3 A }{(sin ^3 A \,- cos^3 A)}$

$\Rightarrow \, \, \, \,LHS \, = \, \dfrac {(sin A cos A \, + \,1 )(sin A \, - \, cos A ) sin^2 A \, cos ^2 A}{(sin A \, - \, cos A)(sin^2A \, + \, cos^2A \, + \, sin A cos A )} \, \, \, \, \, \, [ \because \, \, a^3 \, - \, b^3 \, = \, (a \, - \, b ) (a^2 \, + \, b^2 \, + ab )]$

$\Rightarrow \, \, \, \,LHS \, = \, \dfrac {(sin A \, cos A \, + \,1 ) sin ^2 A \, cos^2 A }{(1 \, + \, sin A \, cos A )} \, = \, sin^2 A \, cos^2 A \, =$ RHS

Prove that:
$\dfrac{\sin \, \theta}{1 \, + \, \cos \, \theta} \, + \, \dfrac{1 \, + \, \cos \, \theta}{\sin \, \theta} \, = \, 2 \, \text{cosec} \, \theta$

We have,

LHS =

$\dfrac{sin\theta}{1 \, + \, cos \, \theta} \, + \, \dfrac{1 \, + \, cos \, \theta}{sin \, \theta}$

$\Rightarrow$ LHS =

$\dfrac{sin^2 \, \theta \, +(1 \, + \, cos \, \theta)^2}{sin \, \theta(1 \, + \, cos \, \theta)}$

$\Rightarrow$ LHS =

$\dfrac{sin^2 \, \theta \, + \, 1 \, + \, 2 \, cos \, \theta \, + \, cos^2 \, \theta}{sin \, \theta(1 \, + \, cos \, \theta)}$

$\Rightarrow$ LHS =

$\dfrac{(sin^2 \, \theta \, + \, cos^2 \, \theta) \, + \, 1 \, + \, 2 \, cos \, \theta}{sin \, \theta(1 \, + \, cos \, \theta)}$

$[\because \, sin^2 \, \theta \, + \, cos^2 \, \theta \, = \, 1]$

$\Rightarrow$ LHS =

$\dfrac{2 \, + \, 2 \, cos \, \theta}{sin \, \theta(1 \, + \, cos \, \theta)} \, = \, \dfrac{2(1 \, + \, cos \, \theta)}{sin \, \theta(1 \, + \, cos \, \theta)} \, = \, \dfrac{2}{sin \, \theta} \, = \, 2 \, cosec \, \theta \, = \, RHS$

Prove the following statements.
$\cos^{6} A + \sin^{6}A = 1 - 3\sin^{2} A + \cos^{2} A$ .

Step 1: Simplify the given expression

LHS

$\Rightarrow$

$\sin^6 A+ \cos^6$ A

$\Rightarrow$

$(\sin^2 A)^3 + (\cos^2 A)^3$

using,

$\, a^3 \, + \, b^3 \, = \, (a \, + \, b)(a^2 \, - \, ab \, + \, b^2)$ , we get

$\Rightarrow$

$(\sin^2 \,A + \, \, \cos^2 \, A) \left\{(\sin^2 \, A)^2 \, + \, (\cos^2 \, A)^2 \, - \, \sin^2 \, A \, \cos^2 \, A) \right\}$

Using $(sin ^2 A +cos^2 A=1)$ , we get

$\Rightarrow$ LHS

$=\left\{(\sin^2 \, A)^2 \, + \, (\cos^2 \, A)^2 \, + \, 2\sin^2 \, A cos^2 A \, -2 \, \sin^2 \, A \, \cos^2 \, A \, - \, \sin^ \, A \, \cos^2 \, A\right\}$

$=\left\{(\sin^2 \, A \, + \, \cos^2 \, A)^2 \, - \, 3\sin^2 \, A \, \cos^2 \, A \, \right\} \, = \, 1 \, - \, 3\sin^2 \, A \, \cos^2 \, A \, = \, RHS$

Hence Proved

Find the number of values of $\theta$ satisfying the equation $\sin 3\theta = 4\sin \theta .\sin 2\theta .\sin 4\theta$ in $0 \le \theta \le 2\pi$

$4\sin { \theta } \sin { 2\theta } \sin { 4\theta } =\sin { 3\theta } \\ \Rightarrow 4\sin { \theta } \sin { (3\theta -\theta ) } \sin { (3\theta +\theta ) } =\sin { 3\theta } \\ \Rightarrow 4[\sin { \theta } (\sin { ^{ 2 }3\theta } -\sin { ^{ 2 }\theta } )]=3\sin { \theta } -4\sin { ^{ 3 }\theta } \\ \Rightarrow 4\sin { \theta } \sin { ^{ 2 }3\theta } -4\sin { ^{ 3 }\theta } =3\sin { \theta } -4\sin { ^{ 3 }\theta } \\ \Rightarrow 4\sin { \theta } \sin { ^{ 2 }3\theta } =3\sin { \theta } \\ \Rightarrow \sin { \theta } (4\sin { ^{ 2 }3\theta } -3)=0\\ \Rightarrow \sin { \theta } =0\quad or\quad 4\sin { ^{ 2 }3\theta } -3=0\\ \Rightarrow Now,\sin { \theta } =0\Rightarrow x=n{ \pi }_{ 1 }\quad n\epsilon z\\ \sin { ^{ 2 }3\theta } =\cfrac { 3 }{ 4 } \\ \sin { ^{ 2 }3\theta } ={ (\cfrac { \sqrt { 3 } }{ 2 } ) }^{ 2 }\\ \sin { ^{ 2 }3\theta } =\sin { ^{ 2 }\cfrac { \pi }{ 3 } } \\ 3\theta =m\pi \pm \cfrac { \pi }{ 3 } ,\quad m\epsilon z\\ x=\cfrac { m\pi }{ 3 } \pm \cfrac { \pi }{ 9 } ,\quad m\epsilon z\\ \therefore x=n\pi \quad or\quad x=\cfrac { m\pi }{ 3 } \pm \cfrac { \pi }{ 9 } ,\quad m,n\epsilon z$

Prove that :
$\dfrac{\cos \, \theta}{1 \, - \, \sin \, \theta} \, + \, \dfrac{\cos \, \theta}{1 \, + \, \sin \, \theta} \, = \, 2 \, \sec \, \theta$

We have,
L.H.S. =

$\dfrac{cos \, \theta}{1 \, + \, sin \, \theta} \, + \, \dfrac{cos \, \theta}{1 \, + \, sin \, \theta}$

$\Rightarrow \, \,$ L.H.S. =

$\dfrac{cos \, \theta (1 \, + \, sin \, \theta) \, + \, cos \, \theta \, (1 \, - \, sin \, \theta)}{(1 \, - \, sin \, \theta)(1 \, + \, sin \, \theta)}$

$\Rightarrow \, \,$ L.H.S. =

$\dfrac{cos \, \theta \, + \, cos \, \theta \, sin \, \theta \, + \, cos \, \theta \, - \, cos \, \theta \, sin \, \theta}{1 \, - \, sin^2 \, \theta}$

$\Rightarrow \, \,$ L.H.S. =

$\dfrac{2 \, cos \, \theta}{cos^2 \, \theta}$

$[\because \, 1 \, - \, sin^2 \, \theta \, = \, cos^2 \, \theta]$

$\Rightarrow \, \,$ L.H.S. =

$\dfrac{2}{cos \, \theta} \, = \, 2 \, sec \, \theta \, = \, R.H.S.$

If $\tan \theta + \sin \theta = m$ and $\tan \theta - \sin \theta = n$ , show that

$m^2-n^2 = 4 \sqrt{ mn}$

Given,

$\tan \theta + \sin \theta = m$ and

$\tan \theta - \sin \theta = n$

we have,

$m+n=2\tan \theta$

$m-n=2\sin \theta$

$m^2-n^2=(m+n)(m-n)$

$\Rightarrow m^2-n^2=4\sin \theta \tan \theta$ ..........(1)

$mn=(\tan \theta + \sin \theta)(\tan \theta - \sin \theta)$

$\Rightarrow mn=\tan ^2\theta -\sin ^2\theta$

now,

$4\sqrt{mn}=4\sqrt{\tan ^2\theta -\sin ^2\theta }$

$=4\sqrt{\dfrac{\sin ^2\theta }{\cos ^2\theta }-\sin ^2\theta}$

$\Rightarrow 4\sqrt{mn}=4\sin \theta \tan \theta$ ..........(2)

equating (1) and (2), we get,

$m^2-n^2 = 4 \sqrt{ mn}$

Hence proved.

Solve $(sec^2 \theta - 1)(1 - cosec^2 \theta) = -1$

$(\sec^{2}\theta-1)(1-\csc^{2}\theta)=(\tan^{2}\theta)(-\cot^{2}\theta)=-1$

( as

$\sec^{2}\theta-\tan^{2}\theta=1 \Rightarrow \sec^{\theta}-1=\tan^{2}\theta$ )

(as

$\csc^{2}\theta-\cot^{2}\theta=1 \Rightarrow 1-\csc^{2}\theta=-\cot^{2}\theta$ )

( as

$\cot^{2}\theta=\cfrac{1}{\tan^{2}\theta}$ )

Find the integrals of the function :
$\dfrac{\sin^{2} x}{1+\cos x}$

1205799_1039908_ans_d73fbc3a3a0f4084beb2d2599a38a799.JPG

Prove that $\dfrac{\sin\theta - \cos \theta +1}{\sin \theta + \cos\theta -1}=\dfrac{1}{\sec\theta - \tan \theta},$ using the identity $\sec^2 \theta= 1+ \tan^2\theta.$

$\cfrac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}$

$=\cfrac{\cos\theta(\tan\theta-1+\sec\theta)}{\cos\theta(\tan\theta+1-\sec\theta)}$

$=\cfrac{\tan\theta+\sec\theta-1}{1-(\sec\theta-\tan\theta)}$

We know that,

$\sec^{2}\theta-\tan^{2}\theta=1$

$\therefore \sec\theta+\tan\theta=\cfrac{1}{\sec\theta-\tan\theta}$

$\cfrac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\cfrac{\cfrac{1}{\sec\theta-\tan\theta}-1}{1-(\sec\theta-\tan\theta)}$

$=\cfrac{1-(\sec\theta-\tan\theta)}{(\sec\theta-\tan\theta)(1-(\sec\theta-\tan\theta)}$

$=\cfrac{1}{(\sec\theta-\tan\theta)}$

Find the value of $\sin \frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{4}\sin\frac{7\pi}{14}\sin \frac{9\pi}{14} \sin\frac{11\pi}{14} \sin \frac{13\pi}{14}=\frac{1}{m}$ .Find $m$

1979480_1028636_ans_461732ee09634208b3c749f9dfa548dd.jpg

Find the value of $\dfrac{\cos A - \sin A+1}{\cos A + \sin A-1}- (\text{cosec} A + \cot A)$

Now,

$\dfrac{\cos A - \sin A+1}{\cos A + \sin A-1}- (\text{cosec} A + \cot A)$

$=\dfrac{(\cos A - \sin A+1)(\cos A+\sin A+1)}{(\cos A + \sin A)^2-(1)^2}- (\text{cosec} A + \cot A)$

$=\dfrac{(\cos A +1)^2-(\sin^2A)}{2\sin A.\cos A}- (\text{cosec} A + \cot A)$

$=\dfrac{(\cos^2 A+2\cos A +1)-(\sin^2A)}{2\sin A.\cos A}- (\text{cosec} A + \cot A)$ [ Since

$\sin^2A+\cos^2A=1$ ]

$=\dfrac{1+\cos 2A+2\cos A}{2\sin A.\cos A}- (\text{cosec} A + \cot A)$ [ Since

$\cos^2A-\sin^2A=\cos 2A$ ]

$=\dfrac{2\cos ^2A+2\cos A}{2\sin A.\cos A}- (\text{cosec} A + \cot A)$ [ Since

$1+\cos 2A=2\cos^2A$ ]

$=\cot A+\text{cosec}A-(\cot A+\text{cosec} A)$

$=0$ .

If $\tan \theta+\dfrac {1}{\tan \theta}=2$ , find the value of $\cot^{2}\theta+\dfrac {1}{\cot^{2}\theta}$ .

$\tan { \theta } +\cfrac { 1 }{ \tan { \theta } } =2$

$\cfrac { 1 }{ \cot { \theta } } +\cot { \theta } =2$

${ \left( \cot { \theta } +\cfrac { 1 }{ \cot { \theta } } \right) }^{ 2 }={ 2 }^{ 2 }$

$\cot ^{ 2 }{ \theta } +\cfrac { 1 }{ \cot ^{ 2 }{ \theta } } +2=4$

$\cot ^{ 2 }{ \theta } +\cfrac { 1 }{ \cot ^{ 2 }{ \theta } } =2$

If $\cos x+\sin x=\dfrac {1}{2}$ , Find the $\tan x$ .

$\cos x+\sin x=1/2$

Divide both sides by

$\cos x$

So,

$1+\tan x=\dfrac { 1 }{ 2\cos x } =\dfrac { secx }{ 2 }$

squaring both sides

$\Rightarrow \quad { \left( 1+\tan x \right) }^{ 2 }=\dfrac { { \sec }^{ 2 }x }{ 4 }$

$\Rightarrow \quad 4\left( 1+{ \tan }^{ 2 }x+2\tan x \right) ={ \sec }^{ 2 }x=\left( 1+{ \tan }^{ 2 }x \right)$

$\Rightarrow \quad 4\left( 1+{ \tan }^{ 2 }x+2\tan x \right) =1+{ \tan }^{ 2 }x$

$\Rightarrow \quad 3{ \tan }^{ 2 }x+8\tan x+3=0$

$\Rightarrow \quad \tan x=\dfrac { -8\pm \sqrt { { 8 }^{ 2 }-4\left( 3 \right) \left( 3 \right) } }{ 2.3 }$

$=\dfrac { -8\pm \sqrt { 64-36 } }{ 6 }$

$=\dfrac { -8\pm \sqrt { 28 } }{ 6 }$

$=\dfrac { -8\pm 2\sqrt { 7 } }{ 6 }$

$=\dfrac { -4\pm \sqrt { 7 } }{ 3 }$

$\tan x=\dfrac { -4\pm \sqrt { 7 } }{ 3 }$

Prove the following identity:
$\sin ^8 x - \cos^8x = (\sin^2x - \cos^2x) (1 - 2 \, \sin^2 x \cos^2 x)$

$\sin ^8x-\cos ^8x=(\sin ^4x-\cos ^4x)(\sin ^4x+\cos ^4x)$

$=(\sin ^2x-\cos ^2x)(\sin ^2x+\cos ^2)((\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x)$ (since,

$(a+b)^2-2ab=a^2+b^2$ )

$=(\sin ^2x-\cos ^2x)(1-2\sin ^2x\cos ^2x)$ (since,

$\sin ^2x+\cos ^2x=1$ )

Therefore,

$\sin ^8x-\cos ^8x=(\sin ^2x-\cos ^2x)(1-2\sin ^2x\cos ^2x)$

Prove that: $\dfrac{{1 - {\mathop{\rm sinAcosA}\nolimits} }}{{\cos A\left( {\sec A - \text{cosec}A} \right)}}.\dfrac{{{{\sin }^2}A - {{\cos }^2}A}}{{{{\sin }^3}A + {{\cos }^3}A}} = \sin A$

Given expression:

$\dfrac{{1 - \sin A\cos A}}{{\cos A(\sec A - \cos ecA)}} \times \dfrac{{{{\sin }^2}A - {{\cos }^2}A}}{{{{\sin }^3}A + {{\cos }^3}A}}$

$= \dfrac{{1 - \sin A\cos A}}{{\cos A(\dfrac{1}{{\cos A}} - \dfrac{1}{{\sin A}})}} \times \dfrac{{(\sin A + \cos A)(\sin A - \cos A)}}{{(\sin A + \cos A)({{\sin }^2}A + {{\cos }^2}A - \sin A\cos A)}}$

$= \dfrac{{1 - \sin A\cos A}}{{\cos A(\dfrac{{\sin A - \cos A}}{{\sin A\cos A}})}} \times \dfrac{{(\sin A - \cos A)}}{{1 - \sin A\cos A}}$

$= \sin A$

If $\sin {x} + \sin {y} + \sin {z} =3$ . Find the value of $\cos {x} + \cos {y} + \cos {z} .$

$\sin x + \sin y + \sin z = 3$

$\sin x=1,\sin y=1,\sin z=1$

So,

$x=y=z=\dfrac{\pi}{2}$

${\mathop{\rm cosx}\nolimits} + \cos y + \cos z = \cos \dfrac{\pi }{2} + \cos \dfrac{\pi }{2} + \cos \dfrac{\pi }{2} = 0$

how to find all possible values of 2cos2x=cos0.

$2\cos{ 2x }=\cos{ 0 }\\ \cos{ 0 }=1\\ 2\cos{ 2x }=1\\ \cos2\theta =2{ \cos }^{ 2 }\theta -1\\ 2(2{ \cos }^{ 2 }x-1)=1\\ 4{ \cos }^{ 2 }x-2=1\\ 4{ \cos }^{ 2 }x=3\\ { \cos }^{ 2 }x=\cfrac { 3 }{ 4 } \\ \cos{ x }=\pm \sqrt { \cfrac { 3 }{ 4 } } \\ \cos\cfrac { \pi }{ 6 } =\cfrac { \sqrt { 3 } }{ 2 }$

Graph is cuttinf at four places so answer is four

1024010_1044975_ans_cae68c5b9d9146df9f11f4e313de6759.png

Prove $\dfrac{1}{\sec x -\tan x} -\dfrac{1}{\cos x} =\dfrac{1}{\cos x}-\dfrac{1}{\sec x+\tan x}$

LHS

$\cfrac{1}{\sec{x}-\tan{x}}$

$-\cfrac{1}{\cos{x}}$

$=\cfrac { 1 }{ \cfrac { 1 }{ \cos { x } }- \cfrac { \sin { x } }{ \cos { x } } } -\cfrac { 1 }{ \cos { x } }$

$=\cfrac { ({ \cos { x } })^{ 2 }-1+\sin { x } }{ ({ \cos { x } })(1-\sin { x) } }$

$\sin^{2}\theta+\cos^{2}\theta=1$

$=\cfrac { ({ \sin { x } })^{ 2 }+\sin { x } }{ ({ \cos { x } })(1-\sin { x) } }$

$=\cfrac { ({ \sin { x } })(1-\sin { x }) }{ ({ \cos { x } })(1-\sin { x) } }=\tan{x}$

RHS

$=-\cfrac{1}{\sec{x}+\tan{x}}$

$+\cfrac{1}{\cos{x}}$

$=-\cfrac { 1 }{ \cfrac { 1 }{ \cos { x } } +\cfrac { \sin { x } }{ \cos { x } } } -\cfrac { 1 }{ \cos { x } }$

$=\cfrac { -({ \cos { x } })^{ 2 }-+1+\sin { x } }{ ({ \cos { x } })(1-\sin { x) } }$

$=\cfrac { ({ \sin { x } })^{ 2 }+\sin { x } }{ ({ \cos { x } })(1+\sin { x) } }$

$=\cfrac { ({ \sin { x } })(1+\sin { x }) }{ ({ \cos { x } })(1+\sin { x) } }=\tan{x}$

LHS=RHS

prove that $\dfrac{{\sec \theta }}{{\sec \theta + \tan \theta }}$ = ${{\sec \theta - \tan \theta }}$

Given,

$\dfrac{\sec \theta}{\sec\theta+\tan \theta}$

$=\dfrac{\dfrac{1}{\cos \theta}}{\dfrac{1}{\cos \theta}+\dfrac{\sin \theta}{\cos \theta}}$

$=\dfrac{1}{1+\sin \theta}$

$=\dfrac{1-\sin \theta}{1-\sin^2\theta}$

$=\dfrac{1-\sin \theta}{\cos \theta}$

$=\sec\theta-\tan \theta$

Find $\left( {\cos \,A + \sec A} \right){^2} + {\left( {\sin A + \text{cosec} A} \right)^2} - {\tan ^2}A - {\cot ^2}A.$

$(\cos A+\sec A)^2+(\sin A+\csc A)^2-\tan^2 A-\cot^2A$

$\Rightarrow \sec A=\cfrac {1}{\cos A}$

$\Rightarrow \csc A=\cfrac {1}{\sin A}$

$\Rightarrow \tan A=\cfrac {\sin A}{\cos A}$

$\Rightarrow \cot A=\cfrac {\cos A}{\sin A}$

$\Rightarrow$ put value in expand:

$\Rightarrow \cos^2 A+\cfrac {1}{\cos ^2A}+2+\sin ^2 A+\cfrac {1}{\sin^2A}+2-\cfrac {\sin^2A}{\cos^2A}-\cfrac {cos^2A}{\sin^2A}$

$\Rightarrow 4+\cos^2A+\sin^2A+\cfrac {1}{\sin^2A}+\cfrac {1}{\cos^2A}-\cfrac {\sin^2A}{\cos^2A}-\cfrac {\cos^2A}{sin^2A}$

$\Rightarrow \therefore \sin^2\theta+\cos^2\theta=1$

$\Rightarrow 4+1+\cfrac {\cos^2A+\sin^2A}{\sin^2A\cos^2A}-\cfrac {\sin^4A+\cos^4A}{\sin^2A\cos^2A}$

$\Rightarrow 5+ \cfrac {1}{\sin^2A\cos^2A}-\cfrac {(\sin^2A)^2+(\cos^2A)^2}{\sin^2A\cos^2A}$

$\Rightarrow \because a^2+b^2=(a+b)^2-2ab$

$\Rightarrow 5+\cfrac {1}{\sin^2A\cos^2A}-\cfrac {(\sin^2A+\cos^2A)^2-2\sin^2A\cos^2A}{\sin2A\cos^2A}$

$\Rightarrow 5+ \cfrac {1}{\sin^2A\cos^2A}-\cfrac {1-2\sin^2A\cos^2A}{\sin^2A\cos^2A}$

$\Rightarrow 5+\cfrac {1-1+2\sin^2A\cos^2A}{\sin^2A\cos^2A}$

$\Rightarrow 5+2=7$

Find $m$ if $\displaystyle \sqrt {\frac{{1 - \sin \theta }}{{1 + \sin \theta }}} + \sqrt {\frac{{1 + \sin \theta }}{{1 - \sin \theta }}} = \frac{{ - m}}{{\cos \theta }},\frac{\pi }{2} < \theta < \pi$ .

$\sqrt {\cfrac {1-\sin\theta}{1+\sin \theta} }+\sqrt {\cfrac {1+\sin \theta}{1-\sin \theta}}$

$\Rightarrow \sqrt {\cfrac {(1-\sin \theta )^2}{1-\sin^2\theta}}+\sqrt{\cfrac {(1+\sin \theta)^2}{1-\sin^2\theta}}$

$\because \sin^2A+\cos^2A=1$

$\because (a+b)(a-b)=a^2-b^2$

$\Rightarrow \left|\cfrac {1-\sin \theta}{\cos \theta}\right|+\left|\cfrac {1+\sin \theta}{\cos \theta}\right|$

$\because \theta\epsilon (\pi/2,\pi)$

So,

$\sin \theta>0$ &

$\cos\theta<0$

$\Rightarrow\cfrac {1-\sin\theta}{-\cos\theta}+\cfrac {1+\sin \theta}{-\cos\theta}$

$\Rightarrow \cfrac {-2}{\cos\theta}=\cfrac {-m}{\cos\theta}$

$\Rightarrow m=-2$

If $\tan \left( {A + B} \right) = x$ and ${\rm{tan}}\left( {A - B} \right){\rm{ = y}}$ find the value of $\tan 2A$ and $\tan 2B$

The number $2A$ can be written as $\left( {\left( {A + B} \right) + \left( {A - B} \right)} \right)$ .

$\tan 2A = \tan \left( {\left( {A + B} \right) + \left( {A - B} \right)} \right)$

$= \dfrac {{\tan \left( {A + B} \right) + \tan \left( {A - B} \right)}}{{1 - \tan \left( {A + B} \right)\tan \left( {A - B} \right)}}$

$= \dfrac {{x + y}}{{1 - xy}}$

The number $2B$ can be written as $\left( {\left( {A + B} \right) - \left( {A - B} \right)} \right)$ .

$\tan 2B = \tan \left( {\left( {A + B} \right) - \left( {A - B} \right)} \right)$

$= \dfrac {{\tan \left( {A + B} \right) - \tan \left( {A - B} \right)}}{{1 + \tan \left( {A + B} \right)\tan \left( {A - B} \right)}}$

$= \dfrac {{x - y}}{{1 + xy}}$

Prove that: $\dfrac{1}{\text{cosec}A- \cot A}-\dfrac{1}{\sin A}=\dfrac{1}{\sin A}-\dfrac{1}{\text{cosec}A+\cot A}$ .

$LHS$

$\cfrac { 1 }{ \cfrac { 1-\cos A }{ \sin A } } \cfrac { 1 }{ \sin A } =\cfrac { \sin A }{ 1-\cos A } -\cfrac { 1 }{ \sin A } \\ =\cfrac { { \sin }^{ 2 }A-1+\cos A }{ \sin A(1-\cos A) } \\ =\cfrac { -{ \cos }^{ 2 }A+\cos A }{ \sin A(1-\cos A) } \\ =\cfrac { \cos A(1-\cos A) }{ \sin A(1-\cos A) } \\ =\cot A$

$RHS$

$\cfrac { 1 }{ \sin A } -\cfrac { 1 }{ cosec A+\cot A } \\ =\cfrac { cosec A+\cot A-\sin A }{ \sin A(cosec A+\cot A) } \\ =\cfrac { 1+\cos A-{ \sin }^{ 2 }A }{ \sin A } \times \cfrac { 1 }{ (1+\cos A) } \\ =\cfrac { \cos A+1-{ \sin }^{ 2 }A }{ \sin A } \times \cfrac { 1 }{ (1+\cos A) } \\ =\cfrac { \cos A+{ \cos }^{ 2 }A }{ \sin A } \times \cfrac { 1 }{ (1+\cos A) } \\ =\cfrac { \cos A(1+\cos A) }{ \sin A } \times \cfrac { 1 }{ (1+\cos A) } \\ =\cot A$

$LHS=RHS$

If $\cos \left(\frac{3\pi}{4}+x\right)- \cos \left(\frac{3\pi}{4}-x\right)= -\sqrt m \sin x$ .Find $m$

$\cos\left( \cfrac { 3\pi }{ 4 } +x \right) -\cos\left( { \cfrac { 3\pi }{ 4 } -x } \right) =-\sqrt { m } \sin x$

$=-2\sin\left( \cfrac { \cfrac { 3\pi }{ 4 } +x+\cfrac { 3\pi }{ 4 } -x }{ 2 } \right) \sin\left( \cfrac { 3\pi }{ 4 } +x-\cfrac { 3\pi }{ 4 } +x \right)$

$=-2\sin\left( \cfrac { 3\pi }{ 2\times 2 } \right) \sin\left( \cfrac { 2x }{ 2 } \right)$

$=-2\times \cfrac { 1 }{ \sqrt { 2 } } \sin x$

$=-\sqrt{2}\sin x$

$\because m=2$

Prove that $\dfrac { 1 }{ \sec { A } -\tan { A } } -\dfrac { 1 }{ \cos { A } } =\dfrac { 1 }{ \cos { A } } =\dfrac { 1 }{ \sec { A } +\tan { A } }$ .

$\dfrac{1}{\sec A- \tan A}-\dfrac{1}{\cos A}=\dfrac{1}{\cos A}-\dfrac{1}{\sec A+ \tan A}$

$LHS \dfrac{1}{\sec A- \tan A}-\dfrac{1}{\cos A}$

$\dfrac{1}{\sec A- \tan A}.\dfrac{\sec A+\tan A}{\sec A+ \tan A}-\dfrac{1}{\cos A}$

$\dfrac{\sec A+ \tan A}{\sec^{2}A- \tan^{2}A}-\dfrac{1}{\cos A}$

$\dfrac{\sec A+ \tan A}{1}-\dfrac{1}{\cos A}$

$= \tan A$

$RHS \dfrac{1}{\cos A}-\dfrac{1}{\sec A+ \tan A}+\dfrac{\sec A- \tan A}{\sec A- \tan A}$

$\dfrac{1}{\cos A}-\dfrac{\sec A- \tan A}{\sec^{2}A- \tan^{2}A}$

$\dfrac{1}{\cos A}- \sec A- \tan A$

$= \tan A$

$\therefore LHS=RHS$

If $A$ is not an integeral multiple of $\pi$ , find k such that $\cos \, A . \cos \, 2A . \cos \, 4A . \cos \, 8A = \dfrac{\sin 16 A}{k \sin A}$ .

We have,

$\cos A .\cos\ 2A. \cos\ 4A. \cos\ 8A$

$=\dfrac{1}{2\sin A}[(2sinAcosA)cos2Acos4Acos8A]$

$=\dfrac{1}{2sinA}[(sin2Acos2Acos4Acos8A)$ ]

$={1}/{4sinA}[(2sin2Acos2A)cos4Acos8A]$

$=1/4sinA(sin4Acos4Acos8A)$

$=1/8sinA[(2sin4Acos4A)cos8A]$

$=1/8sinA(sin8Acos8A)$

$=1/16sinA(2sin8Acos8A)$

$=1/16sinA(sin16A)$

$=\dfrac{\sin 16A}{16\sin A}$

So, the value of

$k$ is

$16$ .

Hence, this is the answer.

If $\sec\phi+\tan\phi=x,$ prove that $\sin\phi=\dfrac{x^2-1}{x^2+1}$ .

$\sec \phi + \tan \phi = x$ ,

To prove :

$\sin \phi = \dfrac{x^{2}-1}{x^{2}+1}$

$LHS$

$\sec \phi+ \tan \phi =\dfrac{1}{\cos \phi } +\dfrac{\sin \phi}{\cos \phi}=\dfrac{1+ \sin \phi}{\cos \phi}$

$\dfrac{1+ \sin \phi}{\cos \phi}= x$

Squaring both sides:

$\dfrac{ (1+ \sin \phi)^{2} }{ \cos^{2} \phi } = x^{2}$

$\Rightarrow \dfrac{ (1+\sin \phi)^{2} }{1- \sin^{2} \phi}= x^{2} \quad (\therefore \cos^{2} \phi =1- \sin^{2} \phi)$

$\Rightarrow \dfrac{(1+ \sin \phi)^{2}}{(1+ \sin \phi)(1- \sin \phi)}= x^{2} (\therefore a^{2}- b^{2}= (a+b)(a-b))$

$\Rightarrow \dfrac{1+ \sin \phi}{1- \sin \phi} =x^{2}$

$\Rightarrow 1+ \sin \phi = x^{2}- x^{2} \sin \phi$

$\Rightarrow \sin \phi (1+x^{2} )=x^{2}-1$

$\Rightarrow \sin \phi =\dfrac{x^{2}-1}{x^{2}+1}$ Hence proved

Prove $\dfrac{{{{\tan }^3}\theta }}{{1 + {{\tan }^2}\theta }} + \dfrac{{{{\cot }^3}\theta }}{{1 + {{\cot }^2}\theta }} = \dfrac{{1 - 2{{\sin }^2}\theta {{\cos }^2}\theta }}{{\sin \theta \cos \theta }}$

$\dfrac{{{{\tan }^3}\theta }}{{1 + {{\tan }^2}\theta }} + \dfrac{{{{\cot }^3}\theta }}{{1 + {{\cot }^2}\theta }}$

$= \dfrac{{{{\tan }^3}\theta }}{{{{\sec }^2}\theta }} + \dfrac{{{{\cot }^3}\theta }}{{\cos e{c^2}\theta }}$

$= \dfrac{{\dfrac{{{{\sin }^3}\theta }}{{{{\cos }^3}\theta }}}}{{\dfrac{1}{{{{\cos }^2}\theta }}}} + \dfrac{{\dfrac{{{{\cos }^3}\theta }}{{{{\sin }^3}\theta }}}}{{\dfrac{1}{{{{\sin }^2}\theta }}}}$

$= \dfrac{{{{\sin }^3}\theta }}{{\cos A}} + \dfrac{{{{\cos }^3}\theta }}{{\sin A}}$

$= \dfrac{{{{\sin }^4}\theta + {{\cos }^4}\theta }}{{\cos \theta \sin \theta }}$

$= \dfrac{{{{\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)}^2} - 2{{\sin }^2}\theta {{\cos }^2}\theta }}{{\sin \theta \cos \theta }}$

$= \dfrac{{1 - 2{{\sin }^2}\theta {{\cos }^2}\theta }}{{\sin \theta \cos \theta }}$

$= R.H.S$

If $7\ \text{cosec}\alpha-3\ \cos\alpha=7$ , prove that $7\ \cot\alpha- 3\ \text{cosec}\ \alpha=3.$

Given

$7 \csc \phi - 3 \cot \phi =7$

$T.P , 7 \cot \phi - 3 \csc \phi =3$

$7 \csc \phi - 3 \cot \phi =7$

$= 7 \csc \phi -7 = 3 \cot \phi$

$7 (\csc \phi -1)= 3 \cot \phi$

$= 7 (\csc \phi-1) (\csc \phi +1)= 3 \cot \phi (\csc \phi +1)$

$=7 (\csc^{2} \phi-1) = 3 \cot \phi (\csc \phi +1)$

$=7 \cot^{2} \phi= 3 \cot \phi (\csc \phi +1)$

$= 7 \cot \phi = 3 (\csc \phi +1)$

$= 7 \cot \phi = 3 \csc \phi +3$

$= 7 \cot \phi -3 \csc \phi =3$

Hence proved

If $\cos^22x - \cos^26x = \sin mx \sin8x$ . Find $m$

$\\ cos^22x-cos^26x\\ =(cos2x+cos6x)(cos2x-cos6x)\\=2cos((\frac{2x+6x}{2}))cos((\frac{2x-6x}{2})) \cdot 2sin(\frac{2x+6x}{2})\cdot sin((\frac{2x-6x}{2}))\\= 4cos4x cos2x sin4x sin2x \\= (2sin4x cos4x)(2sin2x cos2x)\\= sin8x sin4x\\\therefore m=4$

Prove the following statement.
$\sqrt {\dfrac {1 - \sin A}{1 + \sin A}} = \sec A - \tan A$ .

$\sqrt{\cfrac{1+sinA}{1-sinA}}$

$=\sqrt{\cfrac{1+sinA}{1-sinA}\times \cfrac{1+sinA}{1+sinA}}$

$=\sqrt{\cfrac{(1+sinA)^{2}}{1-sin^{2}A}}$

$=\sqrt{\cfrac{(1+sinA)^{2}}{cos^{2}A}}$ ......

$[sin²θ+cos²θ=1]$

$=\cfrac{1+sinA}{cosA}$

$=\cfrac{1}{cosA}+\cfrac{sinA}{cosA}$

Using,

$\sec{\theta}=\dfrac{1}{\cos{\theta}}$ and

$[\dfrac{\sin\theta}{\cos\theta}=\tan\theta]$

$=secA+tanA$

Show that
$(\sec \theta-\tan \theta)^{2}=\dfrac{1-\sin \theta}{1+\sin \theta}$

${\left(\sec{\theta}-\tan{\theta}\right)}^{2}$

$={\left(\dfrac{1}{\cos{\theta}}-\dfrac{\sin{\theta}}{\cos{\theta}}\right)}^{2}$

$=\dfrac{{\left(1-\sin{\theta}\right)}^{2}}{{\cos}^{2}{\theta}}$

since

${\cos}^{2}{\theta}=1-{\sin}^{2}{\theta}$

$=\dfrac{1-\sin{\theta}}{1+\sin{\theta}}$

Hence proved

If $3 \cot A= 4$ check whether $\dfrac{1-\tan^2A}{1+\tan^2A}=\cos^2A - \sin^2A$ or not

$\\\>cotA\>=(\frac{4}{3})\>\therefore\>(\frac{3}{4})\>\\LHS(\frac{1-tan^2A}{1+tan^2A})\\\>=(\frac{1-((\frac{3}{4})^2)}{1+((\frac{3}{4})^2)})\>\\=(\frac{1-((\frac{9}{16}))}{1+((\frac{9}{16}))})\>=(\frac{7}{25})\\\>as\>tanA=(\frac{3}{4})\\\>so\>cosA\>=(\frac{4}{5})\>and\>sinA=\>(\frac{3}{5})\\\>RHS=\>cos^2A-sin^2A\\=((\frac{4}{5})^2-(\frac{3}{5})^2)\\=(\frac{16}{25})-(\frac{9}{25})=(\frac{7}{25})\\\therefore\>LHS=RHS$

If $\tan A + \cot A = m \text{cosec}2A$ .Find $m$

$If\quad \tan { A+\cot { A= } } m\csc { 2A } \\ LHS\\ \cfrac { \sin { A } }{ \cos { A } } +\cfrac { \cos { A } }{ \sin { A } } \\ \cfrac { \sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } }{ \sin { A\cos { A } } } \\ \cfrac { 1 }{ \sin { A\cos { A } } } \\ \cfrac { 2 }{ 2\sin { A\cos { A } } } \\ \cfrac { 2 }{ \sin { 2A } } \\ 2\csc { 2A } \\ \therefore Comparing\quad LHS\& RHS\\ 2\csc { 2A=m\csc { 2A } } \\ \therefore m=2$

Solve for $\alpha$
$\dfrac{\text{cosec}^2 \theta}{\text{cosec} \theta - 1} - \dfrac{\text{cosec}^2 \, \theta}{\text{cosec} \theta + 1} = \alpha \sec^2\theta$

$\dfrac{\csc ^2\theta }{\csc \theta -1}-\dfrac{\csc ^2\theta }{\csc \theta +1}$

$=\dfrac{(\csc \theta +1)\csc ^2\theta-(\csc \theta -1)\csc ^2\theta }{\csc^2 \theta -1}$

$=\dfrac{\csc ^2\theta(\csc \theta +1-\csc \theta +1) }{\csc^2 \theta -1}$

$=\dfrac{2\csc ^2\theta}{\csc^2 \theta -1}$

$=2\sec ^2\theta$

$\Rightarrow 2\sec ^2\theta=\alpha \sec ^2\theta$

$\Rightarrow \alpha =2$

Prove $\left( {1 + \,{{\tan }^2}\,\theta } \right)\,{\sin ^2}\,\theta \, = \,{\tan ^2}\,\theta$

$LHS$

$=(1+\tan ^2 \theta)\sin ^2\theta$

$=(1+\dfrac{\sin ^2 \theta}{\cos ^2\theta})\sin ^2\theta$

$=(\dfrac{\cos ^2\theta+\sin ^2 \theta}{\cos ^2\theta})\ \sin ^2\theta$

As we know that

$\cos ^2\theta+\sin ^2 \theta=1$

Therefore,

$=\dfrac{1}{\cos ^2\theta}\times \sin ^2\theta$

$={[\dfrac{\sin \theta}{\cos \theta}]}^{2}$

since

$\dfrac{\sin \theta}{\cos \theta}=\tan \theta$

$LHS=\tan ^2\theta=RHS$

Hence proved.

given $\cot \theta=1$ Find $\dfrac {(1+\sin \theta) (1-\sin \theta)}{ (1+\cos \theta) (1-\cos \theta)}$

We have

$\cot\theta=1$

$\dfrac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$

$\Rightarrow$

$\dfrac{(1)^2-(\sin\theta)^2}{(1)^2-(\cos\theta)^2}$

$[(a+b)(a-b)=a^2-b^2]$

$\Rightarrow$

$\dfrac{1-\sin^2\theta}{1-\cos^2\theta}$

$\Rightarrow$

$\dfrac{\cos^2\theta}{\sin^2\theta}$

$\Rightarrow$

$cot^2\theta$

$\Rightarrow$

$(1)^2$

$\Rightarrow$

$1$

Convert $3$ radian into degree measure.

It is known that $\pi \;{\rm{radians}} = 180^\circ$ . Then,

$1\;{\rm{radian}} = \cfrac{{180^\circ }}{\pi }$

$3\;{\rm{radians}} = 3 \times \cfrac{{180^\circ }}{\pi }$

$= \cfrac{{540^\circ }}{\pi }$

If $tan\beta =2 sin\alpha sin\gamma cosec(\alpha+\gamma),$ then $cot\alpha, cot\beta cost\gamma$ are in

$tan\beta\>=2sin\alpha\>sin\gamma\>cosec(\alpha\>+\gamma\>)\\sin(\alpha\>+\gamma\>)=2sin\alpha\>sin\>\gamma\>cot\beta\>\\(sin\alpha\>cos\gamma\>+cos\alpha\>sin\alpha\>)\\=2sin\alpha\>sin\>\gamma\>cot\beta\\cot\gamma\>+cot\alpha\>=2cot\beta\>\\Hence\>cot\alpha\>,cot\beta\>and\>cot\gamma\>\>\\\>are\>in\>AP$

Solve $\cot ^{ 2 }{ x } -\tan ^{ 2 }{ x } =4\cot { 2x } \cosec { 2x }$

$\dfrac{cos^{2}x}{sin^{2}x}-\dfrac{sin^{2}x}{cos^{2}x}$

$= \dfrac{cos^{4}x-sin^{4}x}{sin^{2}x.cos^{2}x}$

$= \dfrac{(sin^{2}x+cos^{2}x)(cos^{2}x-sin^{2}x)}{sin^{2}x.cos^{2}x}$

$=\dfrac{cos2x}{\dfrac{1}{4}sin^{2}2x}$

$=4cot2x.cosec^22x$

Show that $cos 2A= cos^4A - sin^4A$

${ \cos }^{ 2 }A={ \cos }^{ 4 }A-{ \sin }^{ 4 }A$

$\text R\text H\text S$

${ \cos }^{ 4 }A-{ \sin }^{ 4 }A$

$=({\cos }^{ 2 }A-{ \sin }^{ 2 }A)({ \cos }^{ 2 }A+{ \sin }^{ 2 }A)$

$=({\cos }^{ 2 }A)(1)$

$={\cos }^{ 2 }A$ (proved)

Hence

$LHS=RHS$

Solve $4\cos ^{ 2 }{ x } +6\sin ^{ 2 }{ x } =5$ .

Given that

$4\cos ^2x+6\sin^2 x=5$

This can be written as

$\Rightarrow 4\cos^2x+4sin^2x+2\sin^2x =5$

$\Rightarrow 4(\cos^2x+sin^2x)+2\sin^2x =5$

$\Rightarrow 4+2\sin^2x=5$

$\Rightarrow 2\sin^2x=1$

$\Rightarrow \sin^2x=\dfrac{1}{2}$

$\Rightarrow \sin x=\dfrac{1}{\sqrt 2}$

$\Rightarrow x=45^0$

Prove that: $\sin^2\left(\dfrac{\pi}{8}+\dfrac{A}{2}\right)-\sin^2\left(\dfrac{\pi}{8}-\dfrac{A}{2}\right)=\dfrac{1}{\sqrt{2}}\sin A$
Find the general solution of the equation.

REF.Image.

OA = x

OB = 180 m

$tan\phi = 4/3$

$sin\phi = 4/5$

$sin\phi = OA/O\epsilon$

$sin\phi = \dfrac{x}{180}$

$4/5 = x/180$

$x = 144$

$144 +10 = 154 m$

$sin (\pi/8 + A/2) - sin^{2}(\pi/8 - \dfrac{A}{2}) = \dfrac{1}{\sqrt{2}} sinA$

$sin(\pi/4) sin(A) = \dfrac{1}{\sqrt{2}}sinA$

$A\epsilon (2n \pi)\epsilon R$

1156166_1069335_ans_fe5b8f7874044fb7807740ee4a63e2d6.jpg

Prove $\cfrac { Sin2A }{ SinA } -\cfrac { Cos2A }{ CosA } =SecA$

1244619_1068842_ans_10ef10830ada41f29c7c1888277a9939.jpg

$\cos\theta -\sin\theta= \sqrt2\sin\theta$ show that $\cos\theta+ \sin\theta= \sqrt2\cos\theta$

$\cos\theta -\sin\theta =\sqrt { 2 } \sin\theta$

${ (\cos\theta -\sin\theta ) }^{ 2 }=2{ \sin }^{ 2 }\theta$

${ \cos }^{ 2 }\theta +{ \sin }^{ 2 }\theta -2\sin\theta \cos\theta =2{ \sin }^{ 2 }\theta$

${ \cos }^{ 2 }\theta -{ \sin }^{ 2 }\theta -2\sin\theta \cos\theta =0$

${ \cos }^{ 2 }\theta -{ \sin }^{ 2 }\theta -2\sin\theta \cos\theta-2{ \cos }^{ 2 }\theta =-2{ \cos }^{ 2 }\theta$

$-{ \cos }^{ 2 }\theta -{ \sin }^{ 2 }\theta -2\sin\theta \cos\theta=-2{ \cos }^{ 2 }\theta$

${ \sin }^{ 2 }\theta +{ \cos }^{ 2 }\theta +2\sin\theta \cos\theta=2{ \cos }^{ 2 }\theta$

${ (\sin\theta +\cos\theta ) }^{ 2 }=2{ \cos }^{ 2 }\theta$

$\sin\theta +\cos\theta =\sqrt { 2 } \cos\theta$

$4\cos\theta -3\sec \theta =2\tan \theta$ .

General soln:

$4\cos\theta-3\sec\theta=2\tan\theta$

We are given

$4\cos\theta-3\sec\theta=2\tan\theta$

$4\cos\theta-\dfrac{3}{\cos\theta}=\dfrac{2\sin\theta}{\cos\theta}$

$4(-1\sin^{2}\theta)-3=2\sin\theta$

$\therefore 4\sin^{2}\theta+2\theta-1=0 ..... (1)$

Now, let compare equation

$(1)$ with

$ax^{2}+bx+c=0$

we get

$a=4$ &

$b=1$ &

$c=-1$

now, Discriminant

$D=b^{2}-4ac$

$=(2)^{2}-4(4)(-1)$

$=4+16$

$D=20>0$

$x=\sin \theta=\dfrac{-b\pm\sqrt{D}}{2a}$

$\sin\theta=\dfrac{-2\pm\sqrt{20}}{2(4)}=\dfrac{-2\pm\sqrt{20}}{8}$

now,

$\sin\theta=\dfrac{-2+\sqrt{20}}{8}$ or

$\sin\theta=\dfrac{-2-\sqrt{20}}{8}$

$\sin\theta=\dfrac{-2+2\sqrt{5}}{8}$ or

$\sin\theta=\dfrac{-2-2\sqrt{5}}{8}$

$\sin\theta=\dfrac{-1+\sqrt{5}}{4}$ or

$\sin\theta=\dfrac{-1-\sqrt{5}}{8}$

here

$\sin\alpha_{1}=\dfrac{-1+\sqrt{5}}{4}, \sin\alpha_{2}=\dfrac{-1-\sqrt{5}}{4}; \left[-\dfrac{\pi}{2}\le \alpha_{1}, \alpha_{2}\le \dfrac{\pi}{2}\right]$

now,

$\alpha_{1}=\dfrac{\pi}{10}$ &

$\alpha_{2}=\dfrac{-3\pi}{10}$

$\therefore \boxed{\theta=n\pi +(-1)^{n}\dfrac{\pi}{10}n\in z}$ or

$\boxed{\theta=n\pi+(-1)^{\pi}\left(\dfrac{-3\pi}{10}\right), n\in z}$

simplify $\dfrac { \sin { 2A } }{ 1+\cos { 2A } } =\tan { A }$

Consider the

$L.H.S$

$\Rightarrow \dfrac{\sin 2A}{1+\cos 2A}$

We know that

$\sin 2A=2\sin A \cos A$

and

$\cos2A=2\cos ^2A-1\space \Rightarrow 1+\cos2A=2\cos^2A$

Substituting the values in the aboe expression, we get

$\Rightarrow \dfrac{\sin 2A}{1+\cos 2A}=\dfrac{2\sin A\cos A}{2\cos ^2 A}$

$\Rightarrow \dfrac{\sin 2A}{1+\cos 2A}=\dfrac{\sin A}{cos A}$

$\Rightarrow \dfrac{\sin 2A}{1+\cos 2A}=\tan A=R.H.S$

Therefore,

$L.H.S=R.H.S$

Hence Proved.

Given $\sin \phi = \dfrac{15}{17}$ , find the value of : $\dfrac{3 - 4 \, \sin^2 \phi}{4 \, \cos^2 \phi - 3}$

$\sin\phi =\dfrac{15}{17}$

$\displaystyle \cos \phi =\dfrac{\sqrt{17^{2}-15^{2}}}{17}=\dfrac{\sqrt{64}}{17}=\dfrac{8}{17}$

$\dfrac{3-4\times \left(\dfrac{15}{17}\right)^{2}}{4\left(\dfrac{8}{17}\right)^{2}-3}$

$\dfrac{3\times 17^{2}-4\times 15^{2}}{4\times 8^{2}-3\times 17^{2}}=\dfrac{867-900}{256-867}=\dfrac{33}{611}$

$\therefore \dfrac{3-4 \sin^{2}\phi }{4 \cos^{2}\phi -3}=\dfrac{33}{611}$

$\sin { \cfrac { \theta }{ 2 } } \sin { \cfrac { 7\theta }{ 2 } } +\sin { \cfrac { 3\theta }{ 2 } } \sin { \cfrac { 11\theta }{ 2 } } =\sin { 2\theta } \sin { 5\theta }$

$\therefore \sin { \cfrac { \theta }{ 2 } } \sin { \cfrac { 7\theta }{ 2 } } +\sin { \cfrac { 3\theta }{ 2 } } \sin { \cfrac { 11\theta }{ 2 } }$

$=\cfrac { 1 }{ 2 } \left[ 2\sin { \cfrac { \left( 4\theta +3\theta \right) }{ 2 } } \sin { \cfrac { \left( 4\theta -3\theta \right) }{ 2 } } +2\sin { \cfrac { \left( 7\theta +4\theta \right) }{ 2 } } \sin { \cfrac { \left( 7\theta -4\theta \right) }{ 2 } } \right]$

$=\cfrac { 1 }{ 2 } \left[ \cos { 3\theta } -\cos { 4\theta } +\cos { 4\theta } -\cos { 7\theta } \right]$

$=\cfrac { 1 }{ 2 } \left[ \cos { 3\theta } -\cos { 7\theta } \right]$

$=\sin { \left( \cfrac { 3\theta +7\theta }{ 2 } \right) } \sin { \left( \cfrac { 7\theta -3\theta }{ 2 } \right) }$

$=\sin { 5\theta } \sin { 2\theta }$

Hence, LHS=RHS

$\sin \theta +\cos \theta =1$ find $\theta$ where $\theta \epsilon(\dfrac{-\pi}{2}, \dfrac{\pi}{2})$

$\sin \theta +\cos \theta =1\\\dfrac{\sin \theta }{\sqrt 2}+\dfrac {\cos \theta }{\sqrt 2}=\dfrac 1{\sqrt 2}\\\sin \theta \cos \dfrac \pi 4+\cos \theta \sin \dfrac \pi 4=\sin \dfrac \pi 4\\\sin \left(\theta +\dfrac\pi 4\right )=\sin \dfrac \pi 4\\\theta =0$

Show that $\dfrac{cosa+sina}{cosa-sina}=tan2a + sec 2a$

$\cfrac { \cos a+\sin a }{ \cos a-\sin a } =\tan 2a+\sec 2a$

${\text {RHS}}$

$\cfrac { 2 \tan a }{ 1-{ \tan }^{ 2 }a } +\cfrac { 1 }{ { \cos }^{ 2 }a }$

$=\cfrac { \cfrac { 2 \sin a }{ \cos a } }{ 1-\cfrac { { \sin }^{ 2 }a }{ { \cos }^{ 2 }a } } +\cfrac { 1 }{ { \cos }^{ 2 }a }$

$=\cfrac { 2 \sin a\cos a }{ { \cos }^{ 2 }a-{ \sin }^{ 2 }a } +\cfrac { 1 }{ { \cos }^{ 2 }a }$

$=\cfrac { 2 \sin a\cos a }{ { \cos }^{ 2 }a-{ \sin }^{ 2 }a } +\cfrac { 1 }{ { \cos }^{ 2 }a-{ \sin }^{ 2 }a }$

$=\cfrac { 2 \sin a\cos a+{ \sin }^{ 2 }a+{ \cos }^{ 2 }a }{ { \cos }^{ 2 }a-{ \sin }^{ 2 }a }$

$=\cfrac { { (\sin a+\cos a) }^{ 2 } }{ (\cos a-\sin a)(\cos a+\sin a) }$

$=\cfrac { \cos a+\sin a }{ \cos a-\sin a }$

Hence proof

Prove: $\tan 3A-\tan 2A-\tan A=\tan A\tan 2A\tan 3A$ .

$LHS=\tan 3 A-\tan 2 A-\tan A=\tan (2 A+A)-(\tan 2 A+\tan A)=\dfrac{\tan 2 A+\tan A}{1-\tan 2 A\tan A}-(\tan 2 A+\tan A)$

$=(\tan 2 A+\tan A)\times \dfrac{1-1+\tan 2 A\tan A}{1-\tan 2 A\tan A}=\dfrac{\tan 2 A+\tan A}{1-\tan 2 A\tan A}\times \tan 2 A\tan A=\tan 3 A\tan 2 A\tan A=RHS$

Hence Proved

Prove: $\tan 20^o+\tan 25^o+\tan 20^o\tan 25^o=1$ .

As we know that

$\tan 45^{\circ}=\tan (20^{\circ}+25^{\circ})=\dfrac{\tan 20^{\circ}+\tan 25^{\circ}}{1-\tan 20^{\circ}\tan 25^{\circ}}$

$\implies \dfrac{\tan 20^{\circ}+\tan 25^{\circ}}{1-\tan 20^{\circ}\tan 25^{\circ}}=1$

$\implies \tan 20^{\circ}+\tan 25^{\circ}=1-\tan 20^{\circ}\tan 25^{\circ}$

$\implies \tan 20^{\circ}+\tan 25^{\circ}+\tan 20^{\circ}\tan 25^{\circ}=1$

Prove : $\sqrt{\dfrac{1 + \sin A}{1 - \sin A }} = \sec A + \tan A$

$\sqrt{\dfrac{1+\sin A}{1-\sin A}}= \sqrt{\dfrac{1+\sin A}{1-\sin A}\times \dfrac{1+\sin A}{1-\sin A}}$

$=\sqrt{\dfrac{(1+\sin A)^{2}}{1-sin^{2}A}}$

$=\sqrt{\dfrac{(1+\sin A)^{2}}{\cos^{2}A}}$

$=\dfrac{1+\sin A}{\cos A}$

$=\sec A+\tan A$

Hence

$\sqrt{\dfrac{1+\sin A}{1-\sin A}}= \sec A+\tan A$

$\dfrac{1 + \cot^2 \theta}{1 + \tan^2 \theta} = \left(\dfrac{1 + \cot \, \theta}{1 + \tan \, \theta} \right)^m$

1980499_1075777_ans_90c6a82a3a23454ab30992baa16e4c85.jpg

Prove that: $\sec ^{ 2 }{ \theta } +\csc ^{ 2 }{ \theta } =\sec ^{ 2 }{ \theta } \csc ^{ 2 }{ \theta }$

$L.H.S={ sec }^{ 2 }\theta +{ cosec }^{ 2 }\theta \\ \quad \quad \quad =\dfrac { 1 }{ { cos }^{ 2 }\theta } +\dfrac { 1 }{ { sin }^{ 2 }\theta } \\ \quad \quad \quad =\dfrac { { sin }^{ 2 }\theta +{ cos }^{ 2 }\theta }{ { { sin }^{ 2 }\theta \cdot cos }^{ 2 }\theta } \\ \quad \quad \quad ={ sec }^{ 2 }\theta \cdot { cosec }^{ 2 }\theta \quad =R.H.S$

$\sec ^{ 2 }{ \theta } -\cfrac { \sin ^{ 2 }{ \theta } -2\sin { 4\theta } }{ 2\cos { 4\theta } -\cos ^{ 2 }{ \theta } } =1$

$\displaystyle \frac{\sin^{2}\theta-2\sin4\theta }{2\cos4\theta-\cos^{2}\theta}$

on simplifying using

$\sin4\theta,\cos4\theta$ formulae in terms of

$\sin^{2}\theta,\cos^{2}\theta$ gives

$\displaystyle =\frac{\sin^{2}\theta}{\cos^{2}\theta}=\tan^{2}\theta$

$\text{sec}^{2}\theta-\tan^{2}\theta=1$

$\therefore$ Hence proved

Prove that: $\frac{1- cos2\theta + sin2\theta}{1+ cos2\theta+ sin 2\theta}=tan\theta$

L.H.S= $(\dfrac{1-cos2\theta +sin2\theta }{1+cos2\theta+sin2 \theta})\\=(\cfrac{1-(1-2sin^2\theta)+2sin\theta cos\theta}{1+(2cos^2 \theta -1)+2sin\theta cos \theta}) \\=(\dfrac{2sin \theta (sin\theta +cos \theta)}{2cos\theta (cos\theta +sin\theta)})\\=tan\theta=RHS$

$Prove\quad that\quad \dfrac { \sin { \theta } -\cos { \theta } +1 }{ \sin { \theta } +\cos { \theta -1 } } =\dfrac { 1 }{ \sec { \theta } -\tan { \theta } }$

$L.H.S=\dfrac { sin\theta -cos\theta +1 }{ sin\theta +cos\theta -1 } \\ \quad \quad \quad =\dfrac { \left( sin\theta -cos\theta +1 \right) \left( sin\theta +cos\theta +1 \right) }{ \left( sin\theta +cos\theta -1 \right) \left( sin\theta +cos\theta +1 \right) } \\ \quad \quad \quad =\dfrac { \left( 1+sin\theta \right) ^{ 2 }-{ cos }^{ 2 }\theta }{ \left( sin\theta +cos\theta \right) ^{ 2 }-1 } \\ \quad \quad \quad =\dfrac { 1+{ sin }^{ 2 }\theta +2sin\theta -{ cos }^{ 2 }\theta }{ 1+2sin\theta cos\theta -1 } \\ \quad \quad \quad =\dfrac { 2{ sin }^{ 2 }\theta +2sin\theta }{ 2sin\theta \cdot cos\theta } \\ \quad \quad \quad =\dfrac { 1+sin\theta }{ cos\theta } \\ \quad \quad \quad =\dfrac { 1 }{ cos\theta } +\dfrac { sin\theta }{ cos\theta } \\ \quad \quad \quad =sec\theta +tan\theta \\ \quad \quad \quad =\dfrac { \left( sec\theta +tan\theta \right) \left( sec\theta -tan\theta \right) }{ \left( sec\theta -tan\theta \right) } \\ \quad \quad \quad =\dfrac { { sec }^{ 2 }\theta +{ tan }^{ 2 }\theta }{ sec\theta -tan\theta } \\ \quad \quad \quad =\dfrac { 1 }{ sec\theta -tan\theta } \quad (Proved)$

Solve $\cfrac { 1+\sin { A } }{ \cos { A } } +\cfrac { \cos { A } }{ 1+\sin { A } } =2\sec { A }$

L.H.S

$=\cfrac { 1+\sin { A } }{ \cos { A } } +\cfrac { \cos { A } }{ 1+\sin { A } }$

$=\cfrac { { \left( 1+\sin { A } \right) }^{ 2 }+\cos ^{ 2 }{ A } }{ \cos { A } \left( 1+\sin { A } \right) }$

$=\cfrac { 1+\sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } +2\sin { A } }{ \cos { A } \left( 1+\sin { A } \right) }$ ............

$[sin²θ+cos²θ=1]$

$=\cfrac { 1+1+2\sin { A } }{ \cos { A } \left( 1+\sin { A } \right) }$

$=\cfrac { 2(1+\sin { A } ) }{ \cos { A } \left( 1+\sin { A } \right) }$

$=\cfrac { 2 }{ \cos { A } } =2\sec { A } =$ R.H.S

Prove $\dfrac{\cos \, A - \sin \, A + 1}{\cos \, A + \sin \, A - 1} = \text{csc} \, A + \cot \, A$ , using the identity $\text{csc}^2 A = 1 + \cot^2 A$

$\dfrac{(\cos A- \sin A + 1)}{(\cos A + \sin A - 1)}$ .

Divide both numerator and the denominator by

$\sin A$ .the sum becomes

$\dfrac{(\cot A - 1 + \csc A)}{(\cot A + 1 -\csc A)}$

$=\dfrac{(\csc A + \cot A - 1)}{(\cot A - \csc A +1)}$

$=\dfrac{\csc A+\cot A-(\csc^2 A- \cot^2 A)}{\cot A-\csc A+1}$

$=\dfrac{(\csc A+cotA)-(\csc A+\cot A)(\csc A-\cot A)}{\cot A-\csc A+1}$

$=\dfrac{(\csc A+cotA){[1-(\csc A-\cot A)}]}{ \cot A - \csc A+1}$

$=\dfrac{(\csc A+cotA)(1-\csc A+\cot A)}{ \cot A - \csc A+1}$

$=\dfrac{(\csc A+\cot A)(\cot A-\csc A+1)}{ \cot A - \csc A+1}$

$=(\csc A+\cot A).\dfrac{\cot A - \csc A+1}{\cot A - \csc A+1}$

$=\csc A+\cot A$ .

Identities used:

$1+cot^2 A=\csc^2 A$

$\csc A=\dfrac{1}{\sin A}$

If $x+y+z=xyz$ , prove that $\cfrac { 2x }{ 1-{ x }^{ 2 } } +\cfrac { 2y }{ 1-{ y }^{ 2 } } +\cfrac { 2z }{ 1-{ z }^{ 2 } } =\cfrac { 2x }{ 1-{ x }^{ 2 } } \cfrac { 2y }{ 1-{ y }^{ 2 } } \cfrac { 2z }{ 1-{ z }^{ 2 } }$

Given

$x+y+z=xyz$

Let

$x=\tan { A } ,y=\tan { B } ,z=\tan { C }$

$\therefore \tan { A } +\tan { B } +\tan { C } =\tan { A } \tan { B } \tan { C }$

$\therefore A+B+C=\Pi$

$2A+2B+2C=2\Pi$

Apply

$\tan$ on both sides

$\tan { \left( 2A+2B+2C \right) } =\tan { \left( 2\Pi \right) }$

$\cfrac { \tan { \left( 2A \right) } +\tan { \left( 2B \right) } +\tan { \left( 2C \right) } -\tan { \left( 2A \right) } \tan { \left( 2B \right) } \tan { \left( 2C \right) } }{ 1-\tan { \left( 2A \right) } \tan { \left( 2B \right) } -\tan { \left( 2B \right) } \tan { \left( 2C \right) } -\tan { \left( 2C \right) } \tan { \left( 2A \right) } } =0$

$\therefore \tan { 2A } +\tan { 2B } +\tan { 2C } =\tan { \left( 2A \right) } \tan { \left( 2B \right) } \tan { \left( 2C \right) }$

$\cfrac { 2\tan { A } }{ 1-\tan ^{ 2 }{ A } } +\cfrac { 2\tan { B } }{ 1-\tan ^{ 2 }{ B } } +\cfrac { 2\tan { C } }{ 1-\tan ^{ 2 }{ C } } =\left( \cfrac { 2\tan { A } }{ 1-\tan ^{ 2 }{ A } } \right) \left( \cfrac { 2\tan { B } }{ 1-\tan ^{ 2 }{ B } } \right) \left( \cfrac { 2\tan { C } }{ 1-\tan ^{ 2 }{ C } } \right)$

$\cfrac { 2x }{ 1-{ x }^{ 2 } } +\cfrac { 2y }{ 1-{ y }^{ 2 } } +\cfrac { 2z }{ 1-{ z }^{ 2 } } =\left( \cfrac { 2x }{ 1-{ x }^{ 2 } } \right) \left( \cfrac { 2y }{ 1-{ y }^{ 2 } } \right) \left( \cfrac { 2z }{ 1-{ z }^{ 2 } } \right)$

$\left( \because x=\tan { A } ,y=\tan { B } ,z=\tan { C } \right)$

If $x=a+b\omega +c{ \omega }^{ 2 },y=a\omega +b{ \omega }^{ 2 }+c$ and $z=a{ \omega }^{ 2 }+b+c{ \omega }$ then find the value of $\cfrac { { x }^{ 2 } }{ yz } +\cfrac { { y }^{ 2 } }{ zx } +\cfrac { { z }^{ 2 } }{ xy }$

$x=a+b\omega +c{ \omega }^{ 2 },y=a\omega +b{ \omega }^{ 2 }+c$

$z=a{ \omega }^{ 2 }+b+c{ \omega }$

$x+y+z=a\left( 1+\omega +{ \omega }^{ 2 } \right) +b\left( \omega +{ \omega }^{ 2 }+1 \right) +c\left( { \omega }^{ 2 }+1+\omega \right)$

$=a(0)+b(0)+c(0)$

$x+y+z=0\quad \left( \because 1+\omega +{ \omega }^{ 2 }=0 \right)$

$x+y+z=0$ , then we know that

${ x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 }=3xyz$

$\cfrac { { x }^{ 2 } }{ yz } +\cfrac { { y }^{ 2 } }{ zx } +\cfrac { { z }^{ 2 } }{ xy } =\cfrac { { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } }{ xyz } =\cfrac { 3xyz }{ xyz } =3$

$\therefore \cfrac { { x }^{ 2 } }{ yz } +\cfrac { { y }^{ 2 } }{ zx } +\cfrac { { z }^{ 2 } }{ xy } =3$

If A+B+C= $\pi$ , then $\ secA \ (cosB \ cosC-sinB \ sinC)$ is equal to__________

Given,

$A+B+C=\pi$

$\therefore B+C=\pi -A$

$secA(cosB\cdot cosC-sinB\cdot sinC)$

$=secA\cdot cos(B+C)$ .................................

$\left (\because\cos (c+d)=\cos c\cdot \cos d-\sin c\cdot \sin d\right )$

$=secA\cdot cos(\pi -A)$

$=secA\cdot (-cosA)$

$=-1$

The answer is

$-1$

Prove $\sin(60^0+\theta ) - \sin (60^0-\theta) = \sin\theta.$

$\sin(60+\theta)-\sin(60-\theta)=\sin\theta$

$60+\theta=x, 60-\theta=y$

$\sin x-\sin y$

$2\cos(\cfrac{x+y}{2})\sin(\cfrac{x-y}{2})$

$=2\cos (\cfrac { 60+\theta +60-\theta }{ 2 } )\sin (\cfrac { 60+\theta -60+\theta }{ 2 } )$

$=2(\cfrac { 1 }{ 2 } )\sin \theta =\sin \theta$

Prove $\dfrac{\sin(A-B)}{\sin A \sin B}+\dfrac{\sin(B-C)}{\sin B \sin C}+ \dfrac{\sin(C - A)}{\sin C \sin A}= 0$

$\cfrac{\sin (A-B)}{\sin A\sin B}+\cfrac{\sin (B-C)}{\sin B\sin C}+\cfrac{\sin (C-A)}{\sin C\sin A}=0$

$=\cfrac{\sin A\cos B-\cos A\sin B}{\sin A\sin B}+\cfrac{\sin B\cos C-\cos B\sin C}{\sin B\sin C}+\cfrac{\sin C\cos A-\cos C\sin A}{\sin C\sin A}$

$=\cot B-\cot A+\cot C-\cot B+\cot A-\cot C$

$=0$ (proved)

Find the value of $\sin{2\theta}$ if $\sin{\theta}=\cfrac{-3}{5}$ and $\pi< \theta< 3\cfrac{\pi}{2}$

Given that

$\theta$ lies in

$Q_3$

So,

$2\pi<\theta<3\pi$

Given that

$\sin \theta =\dfrac{-3}{5}$

Therefore,

$\cos \theta =\dfrac{4}{5}$

$\Rightarrow \sin 2 \theta=2\sin\theta \cos \theta$

$\Rightarrow \sin 2\theta=2(\dfrac{-3}{5})(\dfrac{-4}{5})$

$\Rightarrow \sin 2\theta=\dfrac{24}{25}$

The value of sin ( $37^0$ ) $\times$ cos ( $53^0$ ) =

We know that

$\sin 37^o = 0.6$ and

$\cos53^o = 0.6$

So,

$\sin 37^o \times \cos 53^o = 0.6\times 0.6 = 0.36$

Prove by using vectors that $\sin{(\alpha+\beta)}=\sin{\alpha}\cos{\beta}+\cos{\alpha}\sin{\beta}$

Let

$\vec u=(\sin \alpha, \cos \alpha); \space \space \vec v=(\cos \beta, \sin \beta)$

$\Rightarrow \vec u=[\cos(\dfrac{\pi}{2}-\alpha),\sin(\dfrac{\pi}{2}-\alpha)]$

Also,

$\Rightarrow |\vec u|=|\vec v|=1$

$\Rightarrow \vec u. \vec v=|\vec u||\vec v| \cos(\dfrac{\pi}{2}-\alpha -\beta)$

$\Rightarrow (\sin \alpha \cos \beta +\cos \alpha \sin \beta )=1\times 1\times \cos(\dfrac{\pi}{2}-(\alpha +\beta))$

$\Rightarrow \sin \alpha \cos \beta+\cos \alpha \sin \beta= \sin(\alpha +\beta)$

Thus,

$\Rightarrow \sin (\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta$

Solve : $\cos^6 A - \sin^6 A = \cos \, 2 \, A \left (1 - \dfrac{1}{4} \sin^2 \, 2 \, A\right)$

$\cos ^6A-\sin ^6A$

$=(\cos ^2A)^3-(\sin ^2A)^3$

$=\left [ \cos ^2A-\sin ^2A \right ]\left [ \cos ^4A+\cos ^2A\sin ^2A+\sin ^4A \right ]$

$=\cos 2A\left [ \cos ^4A+\sin ^4A+2\cos ^2A\sin ^2A-\cos ^2A\sin ^2A \right ]$

$=\cos 2A\left [ (\cos ^2A-\sin ^2A)^2- \sin ^2A\cos ^2A\right ]$

$=\cos 2A\left ( 1- \dfrac{4\sin ^2A\cos ^2A}{4} \right )$

$=\cos 2A\left ( 1- \dfrac{(2\sin ^2A\cos ^2A)^2}{4} \right )$

$=\cos 2A\left ( 1- \dfrac{\sin ^2 2A}{4} \right )$

$tan3A - tan 2A - tan A = tan3A tan2AtanA$ .

$tan3A=tan(2A+A)\\tan3A=(\frac{tan2A+tanA}{1-tan2AtanA})\\tan3A-tan3Atan2AtanA=tan2A+tanA\\\therefore tan3A-tan2A-tanA=tan3A\times tan2A\times tanA$

Prove that $\dfrac{{{{\sin }^2}A - {{\sin }^2}B}}{{\sin A\cos A - \sin B\cos B}} = \tan \left( {A + B} \right)$

Consider L.H.S

$\dfrac{\sin ^2A-\sin ^2B}{\sin A \cos A -\sin B \cos B }$

$=\because\left[\cos x=1-2\sin ^2\left(\dfrac{x}{2}\right)\sin \left(\dfrac{x}{2}\right)=\dfrac{1-\cos n}{2}\right]$

$=\dfrac{\left(\dfrac{1-\cos 2A}{2}\right)-\left(\dfrac{1-\cos 2B}{2}\right)}{\dfrac{2\sin A\cos A}{2}-\dfrac{2\sin B \cos B }{2}}$

$[\because\sin (2x)=2\sin x \cos x]$

$=\dfrac{\cos 2B-\cos 2A}{\sin 2A-\sin 2B}$

$\left[\because \cos C-\cos D=2\sin \left(\dfrac{C+D}{2}\right)\sin \left(\dfrac{C-D}{2}\right)\right]$

$=\dfrac{2\sin \left(\dfrac{2B+2A}{2}\right)\sin \left(\dfrac{2B+2A}{2}\right)}{2\sin \left(\dfrac{2A-2B}{2}\right)\cos \left(\dfrac{2B+2A}{2}\right)}$

$\left[\because\sin C- \sin D=2\sin \left(\dfrac{C-D}{2}\right)\cos \left(\dfrac{C+D}{2}\right)\right]$

$=\dfrac{\sin (A+B)}{\cos (A+B)}=\tan (A+B)$ Hence, proved

If $m = \dfrac{\sin \, A}{\sin \,B} , n = \dfrac{\cos \, A}{\cos \, B}$ , then prove that $(m^2 - n^2) \sin^2 B = 1 - n^2$

$(m^2-n^2)$

$=\left ( \dfrac{\sin A}{\sin B} \right )^2-\left ( \dfrac{\cos A}{\cos B} \right )^2$

$= \left ( \dfrac{\sin^2 A}{\sin^2 B} \right )-\left ( \dfrac{\cos^2 A}{\cos^2 B} \right )$

$\Rightarrow (m^2-n^2)\sin^2 B=(\sin^2 A-\cos^2 A\tan^2 B)$

$=1-\cos^2 A-\cos^2 A\tan^2 B$

$=1-\cos^2 A(1+\tan^2 B)$

$=1-\cos^2 A\sec^2 B$

$=1-\dfrac{\cos^2 A}{\cos^2 B}$

$=1-\left (\dfrac{\cos A}{\cos B} \right )^2$

$=1-n^2$

Prove that : $4 - 4 \sin^2 \left (\dfrac{x + y}{2}\right) = (\sin x - \sin y )^2 + (\cos x + \cos y)^2$

R.H.S.

$=\left ( sinx-siny \right )^{2}+\left ( cosx+cosy \right )^{2}$

$=\sin^{2}x+\sin^{2}y-2\sin x\cdot \sin y+\cos^{2}x+\cos^{2}y+2\cos x\cdot \cos y$

$=\sin^{2}x+\cos^{2}x+\sin^{2}y+\cos^{2}y+2\cos x\cdot \cos y-2\sin x\cdot \sin y$

$=1+1+2\cos \left ( x+y \right )$

$=2+2\left [ 1-2sin^{2}\left ( \cfrac{x+y}{2} \right ) \right ]$

$=2+2-4sin^{2}\left ( \cfrac{x+y}{2} \right )$

$=4-4sin^{2}\left ( \cfrac{x+y}{2} \right )$

$=$ L.H.S.

The number of real solutions of $\dfrac { \pi} {2} + \cos^{-1} ( \cos x) = | \tan x | , 0 \le x \le 2 \pi$ is ________.

$\begin{array}{l} \frac { \pi }{ 2 } \to { \cos ^{ -1 } }\left( { \cos x } \right) =\left( { \tan x } \right) \\ in\, \, \left( { 0,\, \, \frac { \pi }{ 2 } } \right) \to \, \, value\, \, exist \\ in\, \, \left( { \frac { \pi }{ 2 } ,\, \, \pi } \right) \to \, \, exist \\ in\, \, \left( { \pi ,\, \, \frac { { 3\pi } }{ 2 } } \right) \to \, \, value\, \, exist \\ in\, \, \left( { \frac { { 3\pi } }{ 2 } ,\, \, 2\pi } \right) \to \, \, value\, \, exist \\ Total\, \, 4\, \, solution. \end{array}$

Find principal and general solution of the equation, $\cot x =-\sqrt{3}$ .

$\cot x=-\sqrt{3}$

We know,

$\cot \dfrac{\pi}{6}=\sqrt{3}$

Therefore,

$\cot (\pi-\dfrac{\pi}{6})=-\cot \dfrac{\pi}{6}=-\sqrt{3}$ and

$\cot(2\pi-\dfrac{\pi}{6})=-\cot \dfrac{\pi}{6}=-\sqrt{3}$

$\cot \dfrac{5\pi}{6}=-\sqrt{3}$ and

$\cot \dfrac{11\pi}{6}=-\sqrt{3}$

Therefore, the principal solutions are

$x=\dfrac{5\pi}{6}$ and

$\dfrac{11\pi}{6}$ .

Now,

$\cot x =\cot \dfrac{5\pi}{6}$

$\tan x =\tan \dfrac{5\pi}{6}$

$x=n\pi+\dfrac{5\pi}{6}$

Therefore, the general solution is

$x=n\pi+\dfrac{5\pi}{6}, n\in Z$ .

Prove that , $\dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{1}{{\sec \theta - \tan \theta}}$

$\dfrac { \sin \theta - \cos \theta + 1 } { \sin \theta + \cos \theta - 1 }$

$= \dfrac { \tan \theta - 1 + \sec \theta } { \tan \theta + 1 - \sec \theta }$

$= \dfrac { ( \tan \theta + \sec \theta ) - 1 } { ( \tan \theta - \sec \theta ) + 1 }$

$= \dfrac { \{ ( \tan \theta + \sec \theta ) - 1 \} ( \tan \theta - \sec \theta ) } { \{ ( \tan \theta - \sec \theta ) + 1 \} ( \tan \theta - \sec \theta ) }$

$= \dfrac { \left( \tan ^ { 2 } \theta - \sec ^ { 2 } \theta \right) - ( \tan \theta - \sec \theta ) } { \{ \tan \theta - \sec \theta + 1 \} ( \tan \theta - \sec \theta ) }$

$= \dfrac { - 1 } { \tan \theta - \sec \theta }$

$= \dfrac { 1 } { \sec \theta - \tan \theta }$

Find the principal solutions of the following equation
$\sin{x}=-\dfrac{1}{2}$

1351860_1130902_ans_3ab263eb199a4e6f870cfa9e1bf08896.jpg

If $cosec\theta +\cot\theta =p$ , then prove that $\cos\theta =\dfrac{p^2-1}{p^2+1}$ .

given

$co\sec { \theta } +\cot { \theta } =P...(1)$

WE know that

$1+\cot ^{ 2 }{ \theta } =co\sec ^{ 2 }{ \theta } \Rightarrow co\sec ^{ 2 }{ \theta } -\cot ^{ 2 }{ \theta } =1$

$\Rightarrow \left( co\sec { \theta } -\cot { \theta } \right) \left( co\sec { \theta } +\cot { \theta } \right) =1\Rightarrow \left( co\sec { \theta } -\cot { \theta } \right) P=1$

$\Rightarrow co\sec { \theta } -\cot { \theta } =\cfrac { 1 }{ P } ...(2)$

Adding (1) and (2)

$o\sec { \theta } =\cfrac { P+\cfrac { 1 }{ P } }{ 2 } ;\cot { \theta } =\cfrac { P-\cfrac { 1 }{ P } }{ 2 }$

$\cos { \theta } =\cfrac { \cot { \theta } }{ co\sec { \theta } } =\cfrac { \cfrac { P-\cfrac { 1 }{ P } }{ 2 } }{ \cfrac { P+\cfrac { 1 }{ P } }{ 2 } } =\cfrac { { P }^{ 2 }-1 }{ { P }^{ 2 }+1 }$

Hencd proved

Find all solution of the equation.
$\sin x+\sin \dfrac{\pi}{8}\sqrt{(1-\cos x)^2+\sin^2x}=0$ in $\left[\dfrac{5\pi}{2}, \dfrac{7\pi}{2}\right]$ .

$\begin{matrix} \sin x+\sin \frac { \pi }{ 8 } \sqrt { { { (1-\cos x) }^{ 2 } }+{ { \sin }^{ 2 } }x } =0 \\ \Rightarrow \sin x+\sin \frac { \pi }{ 8 } \sqrt { 1-2\cos x+{ { \cos }^{ 2 } }x+{ { \sin }^{ 2 } }x } =0 \\ \Rightarrow \sin x+\sin \frac { \pi }{ 8 } \sqrt { 2\left( { 1-\cos x } \right) } =0 \\ \Rightarrow \sin x+2\sin \frac { \pi }{ 8 } \sqrt { { { \sin }^{ 2 } }\frac { x }{ 2 } } =0 \\ \Rightarrow 2\sin \frac { x }{ 2 } \cdot \cos \frac { x }{ 2 } +2\sin \frac { x }{ 2 } \cdot \sin \frac { \pi }{ 8 } =0 \\ \Rightarrow 2\sin \frac { x }{ 2 } \left[ { \cos \frac { x }{ 2 } +\sin \frac { \pi }{ 8 } } \right] =0 \\ \therefore \sin \frac { x }{ 2 } =0 \\ \frac { x }{ 2 } =n\pi \\ x=2n\pi ,n\in I \\ putn=1,2,3 \\ x=2\pi ,3\pi \in \left[ { \frac { { 5\pi } }{ 2 } ,\frac { { 7\pi } }{ 2 } } \right] \\ \end{matrix}$

Prove the following identity :
$\frac{1}{sin{\theta}+cos{\theta}}+\frac{1}{sin{\theta}-cos{\theta}}=\frac{2sin{\theta}}{1-2cos^2{\theta}}$

$\frac { 1 } { \sin \theta + \cos \theta } + \frac { 1 } { \sin \theta - \cos \theta }$

$\begin{aligned} & = \frac { \sin \theta - \cos \theta + \sin \theta + \cos \theta } { ( \sin \theta + \cos \theta ) ( \sin \theta - \cos \theta ) } \\ & = \frac { 2 \sin \theta } { \sin ^ { 2 } \theta - \cos ^ { 2 } \theta } \\ = & \frac { 2 \sin \theta } { 1 - \cos ^ { 2 } \theta - \cos ^ { 2 } \theta } = \frac { 2 \sin \theta } { 1 - 2 \cos ^ { 2 } \theta } \end{aligned}$

If $m = \left( {\cos \theta - \sin \theta } \right)$ and $n = \left( {\cos \theta + \sin \theta } \right)$ then show that $\sqrt {{m \over n}} + \sqrt {{n \over m}} = {2 \over {\sqrt {1 - {{\tan }^2}\theta } }}$

$m=\cos \theta-\sin\theta$

$n=\cos\theta\sin\theta$

$\sqrt{\dfrac{m}{n}}+\sqrt{\dfrac{n}{m}}=\dfrac{m+n}{\sqrt{mn}}$

$m+n=\cos \theta-\sin\theta+\cos\theta+\sin\theta=2\cos\theta$

$\sqrt{mn}=\sqrt{(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)}$

$=\sqrt{\cos^{2}\theta-\sin^{2}\theta}$

$=\sqrt{\cos 2\theta}$

$\dfrac{m+n}{\sqrt{mn}}=\dfrac{2\cos\theta}{\sqrt{\cos 2\theta}}=2\sqrt{\dfrac{\cos^{2}\theta}{\cos^{2}\theta-\sin^{2}\theta}}$

$=2\sqrt{\dfrac{1}{\dfrac{\cos^{2}\theta-\sin^{2}\theta}{\cos^{2}\theta}}}$

$\sqrt{\dfrac{m}{n}}+\dfrac{n}{m}=\dfrac{2}{\sqrt{1-\tan^{2}\theta}}$

Evaluate :
$\cos (40^o + \theta) - \sin(50^o - \theta) + \cos 240^o + \cos 250^o \sin 240^o + \sin 250^o=?$

Use concept

$\cos(90^o-\theta)=\sin\theta$

and

$\sin(90^o-\theta)=\cos\theta$

Then

$\cos(40^o+\theta)-\sin(50^o-\theta)+\dfrac{\cos^240^o+\cos^250^o}{\sin^240^o+\sin^250^o}$

$=\cos(40^o+\theta)-\sin(50^o-\theta)+\dfrac{\cos^2(90^o-50^o)+\cos^2(90^o-40^o)}{\sin^240^o+\sin^250^o}$

$=\cos(40^o+\theta)-\sin(50^o-\theta)+\dfrac{\sin^250^o+\sin^240^o}{\sin^240^o+\sin^250^o}$

$=\cos(40^o+\theta)-\sin[90^o-(40^o+\theta)]+1$

$=\cos(40^o+\theta)-\cos(40^o+\theta)+1=0+1=1$ .

If ${\tan ^{ - 1}}\alpha + {\tan ^{ - 1}}\beta = \dfrac{\pi }{4}$ , then twice write the value of $\alpha + \beta + \alpha \beta .$

$tan^{-1}(\alpha )+tan^{-1}(\beta ) = \dfrac{\pi }{4}$

$\Rightarrow tan^{-1}(\dfrac{\alpha +\beta }{1-\alpha \beta }) = \dfrac{\pi }{4}$

$\Rightarrow \dfrac{\alpha +\beta }{1-\alpha \beta } = tan(\dfrac{\pi }{4}) = 1$

$\Rightarrow \alpha +\beta = 1-\alpha \beta$

$\Rightarrow \alpha +\beta +\alpha \beta = 1$

$\therefore \alpha +\beta +\alpha \beta = 1$ and

$2(\alpha +\beta +\alpha \beta ) = 2$

1107387_1134772_ans_16dab2ec0b92454cb6fa2e372e907ef4.jpg

Find the principal solutions of the following equation
$\tan{x}=-1$

1351859_1130908_ans_107dc264a9cc44bf823962f0a3ab54ca.jpg

If $\sqrt{3} \sin \, \theta = \cos \, \theta$ , find the value of $\dfrac{\sin \theta \, \, \tan \theta ( 1 + \cos \, \theta)}{\sin \theta + \cos \theta}$

$\sqrt{3}\sin \theta =\cos \theta$

$\dfrac{\sin \theta}{\cos \theta }=\dfrac{1}{\sqrt{3}}$

$\Rightarrow \tan \theta =\dfrac{1}{\sqrt{3}}$

use pythagorean triplets:

$1,\sqrt{2},\sqrt{3}$

$\tan \theta =\dfrac{1}{\sqrt{3}}\Rightarrow \sin \theta =\dfrac{1}{\sqrt{2}},\cos \theta =\dfrac{\sqrt{3}}{\sqrt{2}}$

$\dfrac{\sin \theta \tan \theta(1+\cos \theta ) }{\sin \theta+\cos \theta }$

$=\dfrac{\dfrac{1}{\sqrt{2}}\dfrac{1}{\sqrt{3}}\left ( 1+\dfrac{\sqrt{3}}{\sqrt{2}} \right )}{\dfrac{1}{\sqrt{2}}+\dfrac{\sqrt{3}}{\sqrt{2}}}$

$=\dfrac{\dfrac{1}{\sqrt{6}}\left ( \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{2}} \right )}{\dfrac{1+\sqrt{3}}{\sqrt{2}}}$

$=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{6 }\times \sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{3 }+1}$

$=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{6 }( \sqrt{3}+1)}$

Find the principal solutions of the following equation
$\sqrt{3}\csc{x}+2=0$

1351856_1130914_ans_0c4c5dec35ed4109a7a5f5987e8bce74.jpg

Find the principal solutions of the following equation
$\tan{x}=-\sqrt{3}$

$\tan{x}=\tan{a}$

$x=n(\pi)+a$

$\tan{x}= \tan(-60)$

$x=n(\pi)-60$

If $\cos \theta - \sin \theta = 1$ , show that $\cos \theta + \sin \theta = 1$

Given,

$\cos\theta -\sin\theta =1$

Squaring on both sides

$(\cos\theta -\sin\theta)^2=1$

$\cos^2\theta +\sin^2\theta -2\cos\theta \sin\theta =1$

$1-2\cos\theta \sin\theta =1$

$2\cos\theta \sin\theta =0$

$(\cos \theta +\sin\theta)^2$

$=\cos^2\theta +\sin^2\theta +2\sin\theta \cos\theta$

$(\cos\theta +\sin\theta)^2=1+0$

$\therefore \cos\theta +\sin\theta =1$ .

Find the principal solution of the following equation:
$\sin{x}=\frac { \sqrt { 3 } }{ 2 }$

1972424_1147257_ans_570170f5702f4ac182c30f54ead4d0a1.jpeg

Find the principal solution of the following equation:
$\cos{x}=\frac{1}{2}$

1972422_1147253_ans_f8ee81f8f14b4fbb9b127bc79db66fbb.jpeg

Evaluate
$\cos^{4}\theta+\sin^{4}\theta=1-2\cos^{2}\theta.\sin^{2}\theta$

$R.H.S= 1- 2\cos^2\theta . \sin^2\theta$

$=[\sin^2\theta+\cos^2\theta]^2 -2\cos^2\theta . \sin^2\theta$

$= \sin^4\theta + \cos^4 \theta +2\cos^2\theta . \sin^2\theta -2\cos^2\theta . \sin^2\theta$

$=\sin^4\theta + \cos^4 \theta$

Hence proved

If $\cot \theta + \tan \theta = x$ and $\sec \theta - \cos \theta = y$ , then prove that $x^2 y \left(\dfrac{2}{3}\right) xy^2 \left(\dfrac{2}{3}\right) = 1$

The question should be:

prove that

$(x^2y)^{\frac{2}{3}}-(xy^2)^{\frac{2}{3}}=1$

$\cot \theta +\tan \theta =\dfrac{1+\tan ^2\theta }{\tan \theta }=\dfrac{\sec ^2\theta }{\tan \theta}$

$\Rightarrow x=\cot \theta +\tan \theta =\dfrac{1}{\sin \theta \cos \theta }$

$\sec \theta -\cos \theta =\dfrac{1\cos ^2\theta }{\cos \theta }=\dfrac{\sin ^2\theta }{\cos \theta}$

$\Rightarrow y=\sec \theta -\cos \theta =\dfrac{\sin ^2\theta }{\cos \theta}$

$\therefore x^2=\dfrac{1}{\sin ^2\theta \cos ^2\theta },y^2=\dfrac{\sin ^4\theta }{\cos ^2\theta}$

$\therefore x^2y=\dfrac{1}{\sin ^2\theta \cos ^2\theta }\times \dfrac{\sin ^2\theta }{\cos \theta}=\sec ^3\theta$

$\therefore xy^2=\dfrac{1}{\sin \theta \cos \theta }\times \dfrac{\sin ^4\theta }{\cos ^2\theta}=\tan ^3\theta$

$\therefore (x^2y)^{\frac{2}{3}}-(xy^2)^{\frac{2}{3}}=\left ( \sec ^3\theta \right )^{\frac{2}{3}}-\left ( \tan ^3\theta \right )^{\frac{2}{3}}$

$=\sec ^2\theta -\tan ^2\theta$

$=1$

Evaluate: $\frac{{{{\cos }^2}\left( {{{45}^ \circ } + \theta } \right) + {{\cos }^2}\left( {{{45}^ \circ } - \theta } \right)}}{{\tan \left( {{{60}^ \circ } + \theta } \right)\tan \left( {{{30}^ \circ } - \theta } \right)}} + \cos ec\left( {{{75}^ \circ } + \theta } \right) - \sec \left( {{{15}^ \circ } - \theta } \right)$

$\dfrac{{\cos}^{2}{\left(45+\theta\right)}+{\cos}^{2}{\left(45-\theta\right)}}{\tan{\left(60+\theta\right)}\tan{\left(30-\theta\right)}}+\csc{\left(75+\theta\right)}-\sec{\left(15-\theta\right)}$

$=\dfrac{{\cos}^{2}{\left(45+\theta\right)}+{\cos}^{2}{\left(90-\left(45+\theta\right)\right)}}{\tan{\left(60+\theta\right)}\tan{\left(90-\left(60+\theta\right)\right)}}+\csc{\left(90-\left(15-\theta\right)\right)}-\sec{\left(15-\theta\right)}$

$=\dfrac{{\cos}^{2}{\left(45+\theta\right)}+{\sin}^{2}{\left(45+\theta\right)}}{\tan{\left(60+\theta\right)}\cot{\left(60+\theta\right)}}+\sec{\left(15-\theta\right)}-\sec{\left(15-\theta\right)}$

$=\dfrac{1}{1}+0=1$ since

${\sin}^{2}{\theta}+{\cos}^{2}{\theta}=1$

$\dfrac{{\cos \,{{135}^ \circ } - \cos \,{{120}^ \circ }}}{{\cos \,{{135}^ \circ } + \cos \,{{120}^ \circ }}} = (3 - 2\sqrt 2 )$

$\dfrac{\cos 135^{\circ}- \cos 120^{\circ}}{\cos 135^{\circ}+\cos 120^{\circ}}=\dfrac{\cos(180-45)- \cos (90+30)}{\cos (180-45)+\cos (90+30)}$

$=\dfrac{-\cos 45^{\circ}+ \sin 30^{\circ}}{- \cos 45^{\circ}- \sin 30^{\circ}}$

$= \dfrac{-\dfrac{1}{\sqrt{2}}+\dfrac{1}{2}}{-\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}} =\dfrac{-2+\sqrt{2}}{-2-\sqrt{2}}\times \dfrac{-2+\sqrt{2}}{-2+\sqrt{2}}$

$=\dfrac{(-2+\sqrt{2})^{2}}{4-2}=3-2\sqrt{2}$

solve $2|\sin x| \geqslant |\,\,x\varepsilon \,[0,4\pi ]$

$\Rightarrow 2|\sin x|\ge 1$

$x\in[0, \pi]$

The above inequality can be written as,

$\Rightarrow 2\sin \ge 1$ or

$\Rightarrow -2\sin x\ge 1$

$\Rightarrow \sin \ge \dfrac{1}{2}$

$\Rightarrow \sin x\le \dfrac{-1}{2}$

$\Rightarrow x\ge \sin^{-1}\dfrac{1}{2}$

$\Rightarrow x\le \sin^{-1}\dfrac{-1}{2}$

$\Rightarrow x\ge \dfrac{\pi}{6}$

$\Rightarrow x\le \dfrac{-\pi}{6}$

(Not possible as

$x\in[0, 4, \pi]$ given)

$\therefore x\ge\dfrac{\pi}{6}$

if $\sin \theta = - \dfrac{4}{5},\,\pi < \theta < \dfrac{{3\pi }}{2},$ then find
$\sin 2\theta$
$\cos 2\theta \,$
$\tan 2\theta \,$

Solution :-

Given

$sin \theta = \dfrac{-4}{5}$ and

$\pi < \theta < \dfrac{3\pi }{2}$

then

$1.sin2\theta = 2sin\theta .cos\theta$

$cos \theta = -\dfrac{\sqrt{5^{2}-4^{2}}}{5}=-\dfrac{3}{5}$

$\because \pi < \theta < \dfrac{3\pi }{2}$

$\therefore sin 2\theta =2\times \dfrac{-4}{5}\times \dfrac{-3}{5}= \dfrac{24}{25}$

$2.cos 2\theta = cos^{2}\theta - sin^{2}\theta$

$=(\dfrac{-3}{5})^{2}-(\dfrac{-4}{5})^{2}$

$= \dfrac{9}{25}-\dfrac{16}{25}$

$= \dfrac{-7}{25}$

$3. tan 2\theta = \dfrac{sin2\theta }{tan 2\theta }=\dfrac{24/25}{-7/25}$

$=\dfrac{-24}{7}$

1107551_1165765_ans_41f88b55a7b74767997287226aeab504.jpg

Find value of
$\sin^2A{\,\,}\cot^2A +\cos^2A{\,\,}\tan^2{\,\,}A$

$\sin^2A\cot^2A +\cos^2A\tan^2A=$

$\sin^2A{\,\,}\dfrac{\cos^2A}{\sin^2A} +\cos^2A\dfrac{\sin^2A}{\cos^2A}=\cos^2A+\sin^2A=1$

If $\cos \theta + \sin \theta = \sqrt{2} \cos \theta$ , then prove that $\cos \theta - \sin \theta = \sqrt{2} \sin \theta$

1044887_1176131_ans_a7af2c97ce894185bad467bdef8b388c.png

Prove that $\sec^{4}\theta-\sec^{2}\theta=\tan^{4}\theta+\tan^{2}\theta$

$\sec^4\theta -\sec^2\theta =\tan^4\theta +\tan^2\theta$

$LHS$

$\sec ^4-\sec^2\theta$

$\Rightarrow \ \sec^2\theta (\sec^2\theta -1)$

$\Rightarrow \ \sec^2\theta (\tan^2\theta )$

$\Rightarrow \ (1+\tan^2 \theta )(\tan^2 \theta )$

$\Rightarrow \ \tan^2 \theta +\tan^4 \theta$

$\Rightarrow \ RHS$

$\Rightarrow \ LHS=RHS$

Solve the trigonometric equation
$\dfrac{(\cot\ A-1+\text{cosec}\ A)^{2}}{(\cot\ A)^{2}-(1-\text{cosec}\ A)^{2}}$

$\dfrac{{\left(\cot{A}-1+\csc{A}\right)}^{2}}{{\cot}^{2}{A}-{\left(1-\csc{A}\right)}^{2}}$

$=\dfrac{{\left(\cot{A}-1+\csc{A}\right)}^{2}}{\left(\cot{A}-1+\csc{A}\right)\left(\cot{A}+1-\csc{A}\right)}$

$=\dfrac{\left(\cot{A}-1+\csc{A}\right)}{\left(\cot{A}-\csc{A}+1\right)}$

$=\dfrac{\left(\cot{A}+\csc{A}-1\right)}{\left(\cot{A}-\csc{A}+1\right)}$

$=\dfrac{\left(\cot{A}+\csc{A}-\left({\cot}^{2}{A}-{\csc}^{2}{A}\right)\right)}{\left(\cot{A}-\csc{A}+1\right)}$

$=\dfrac{\left(\cot{A}+\csc{A}\right)\left(1-\csc{A}+\cot{A}\right)}{\left(\cot{A}-\csc{A}+1\right)}$

$=\cot{A}+\csc{A}$

$=\dfrac{\cos{A}+1}{\sin{A}}$

$=\dfrac{2{\cos}^{2}{\dfrac{A}{2}}-1+1}{2\sin{\dfrac{A}{2}}\cos{\dfrac{A}{2}}}$

$=\dfrac{\cos{\dfrac{A}{2}}}{\sin{\dfrac{A}{2}}}$

$=\cot{\dfrac{A}{2}}$

Solve:- $\dfrac{{x\sin x}}{{1 + \cos x}}=0$

1358577_1174580_ans_02334ffaeaa744b0bd8aafab1c0013ac.jpg

Find the principal solution of $\cos x = - \frac{1}{2}$

$\cos x = -1/2$

$\Rightarrow -\dfrac{1}{2}=- \cos \dfrac{\pi}{3}= \cos \left( \pi - \dfrac{\pi}{3} \right) = \cos \dfrac{2 \pi}{3}$

$\Rightarrow x= \dfrac{2 \pi}{3}$

$x \in [0 , \pi]$

The number of solution of the equation, $\sin ^ { 4 } x - \cos ^ { 2 } x \cdot \sin x + 2 \sin ^ { 2 } x + \sin x = 0,0 \leq x \leq 3 \pi$ is ..

${\sin}^{4}{x}-{\cos}^{2}{x}\sin{x}+2{\sin}^{2}{x}+\sin{x}=0$

$\Rightarrow\,{\sin}^{4}{x}+2{\sin}^{2}{x}-{\cos}^{2}{x}\sin{x}+\sin{x}=0$

$\Rightarrow\,{\sin}^{2}{x}\left({\sin}^{2}{x}+2\right)+\sin{x}\left(1-{\cos}^{2}{x}\right)=0$

$\Rightarrow\,{\sin}^{2}{x}\left({\sin}^{2}{x}+2\right)+\sin{x}{\sin}^{2}{x}=0$

$\Rightarrow\,{\sin}^{2}{x}\left({\sin}^{2}{x}+2+\sin{x}\right)=0$

$\Rightarrow\,{\sin}^{2}{x}=0,\,{\sin}^{2}{x}+\sin{x}+2=0$

$\Rightarrow\,\sin{x}=0,\,\sin{x}=\dfrac{-1\pm\sqrt{1-8}}{2}\notin\left[-1,1\right]$

Hence

$\sin{x}=0$

$\Rightarrow\,x=n\pi$

$\therefore\,x=\left\{0,\pi,2\pi,3\pi\right\}$

Number of solutions

$=4$

Solve: $\dfrac{cos7A}{secA}-\dfrac{sin7A}{cosec A}=cos 8A$

We know that,

$\cos(A+B)=\cos A\cos B-\sin A\sin B$

$\dfrac{cos7A}{secA}-\dfrac{sin7A}{cosec A}=\cos7A\cos A-\sin7A\sin A=\cos (7A+A)=\cos8A$

Find the principal solutions of $\cot{x}=-\sqrt{3}$

Given

$\cot x=-\sqrt3$

$\tan x=\dfrac{1}{\cot x}$

$\tan x=\dfrac{1}{-\sqrt3}$

We know that

$\tan 30=\dfrac{1}{\sqrt3}$

We find the value of x where tan is negative

tan is negative in 2nd and 4th quadrant

Value in 2nd quadrant

$=180-30=150^{o}$

Value in 4th quadrant

$=360-30=330^{o}$

So. principal solution are

$x=150^{o}$

$x=150\times\dfrac{\pi}{180}$

$x=\dfrac{5\pi}{6}$

$x=330^{o}$

$x=330\times\dfrac{\pi}{180}$

$x=\dfrac{11\pi}{6}$

The value of $6\left( {{{\sin }^6}\theta + {{\cos }^6}\theta } \right) - 9\left( {{{\sin }^4}\theta + {{\cos }^4}\theta } \right) + 4$ is

$6\left( {{{\sin }^6}\theta + {{\cos }^6}\theta } \right) - 9\left( {{{\sin }^4}\theta + {{\cos }^4}\theta } \right) + 4\\=6(\sin^2\theta+\cos^2\theta)(\sin^4\theta+\cos^4\theta-\sin^2\theta\cos^2\theta)-9(\sin^4\theta+\cos^4\theta)+4\\=6\sin^4\theta+6\cos^4\theta-6\sin^2\theta\cos^2\theta-9\sin^4\theta-9\cos^4\theta+4\\=-3(\sin^4\theta+\cos^4\theta+2\sin^2\theta\cos^2\theta)+4\\=-3(\sin^2\theta+\cos^2\theta)^2+4=-3+4=1$

Solve: $\tan 2 x-\tan 3x=0$

$\tan2x-\tan3x=0\\ \Rightarrow \dfrac { 2\tan x }{ 1-{ \tan }^{ 2 }x } -\dfrac { 3\tan x-{ \tan }^{ 3 }x }{ 1-3{ \tan }^{ 2 }x } =0\\ Let\quad \tan x=a\\ \Rightarrow \dfrac { 2a }{ 1-{ a }^{ 2 } } -\dfrac { 3a-{ a }^{ 3 } }{ 1-3{ a }^{ 2 } } =0\\ \Rightarrow { a }^{ 4 }+2{ a }^{ 2 }+1=0\\ \Rightarrow { \left( { a }^{ 2 }+1 \right) }^{ 2 }=0\\ \Rightarrow { a }^{ 2 }+1=0\\ \Rightarrow { a }^{ 2 }=-1\\$

Prove that $\displaystyle \dfrac { \tan \theta } { 1 - \cot \theta } + \dfrac { \cot \theta } { 1 - \tan \theta } = 1 + \sec \theta \>cosec \theta$

$\\(\dfrac{tan\theta}{1-cot\theta})+(\dfrac{cot\theta}{1-tan\theta})\\=(\dfrac{(\dfrac{sin\theta}{cos\theta})}{1-(\dfrac{cos\theta}{sin\theta})})+(\dfrac{(\dfrac{cos\theta}{sin\theta})}{1-(\dfrac{sin\theta}{cos\theta})})\\=(\dfrac{sin\theta}{cos\theta})\times(\dfrac{sin\theta}{(sin\theta-cos\theta)})+(\dfrac{cos\theta}{sin\theta})\times(\dfrac{cos\theta}{(cos\theta-sin\theta)})\\=(\dfrac{1}{(sin\theta-cos\theta)})\left[(\dfrac{sin^2\theta}{cos\theta})-(\dfrac{cos^2\theta}{sin\theta}) \right ]\\=(\dfrac{1}{(sin\theta-cos\theta)})(\dfrac{sin^3\theta-cos^3\theta}{sin\theta\>cos\theta})\\=(\dfrac{1}{(sin\theta-cos\theta)})(\dfrac{(sin\theta-cos\theta)(sin^2\theta+sin\theta\cos\theta+cos^2\theta)}{sin\theta\cos\theta})\\=(\dfrac{1+sin\theta\cos\theta}{sin\theta\cos\theta})=sec\theta\>cosec\theta+1$

If $\cos \theta > 0, \tan \theta+ \sin \theta=m$ and $\tan \theta -\sin \theta =n$ , then show that $m^{2}-n^{2}=4\sqrt{mn}$ .

$\tan\theta +\sin \theta=m\;\&\;\tan\theta-\sin\theta=n$

$\Rightarrow mn=\left( \tan\theta+\sin\theta\right) \left( \tan\theta -\sin\theta\right)$

$=\tan^2\theta-\sin^2\theta$

$=\sin^2\theta\left( \sec^2\theta-1\right)$

$=\sin^2\theta\times \tan^2\theta$

$\Rightarrow m^2-n^2$

$=\left(m+n\right)\left(m-n\right)$

$=\left( \tan \theta +\sin\theta +\tan\theta-\sin\theta\right) \left( \tan\theta+\sin\theta-\tan\theta+\sin\theta\right)$

$=4\tan\theta\sin\theta$

$=4\sqrt{mn}$

Hence, proved.

The set of angles between $0$ & $2\pi$ satisfying the equation $4{\cos ^2}\theta - 2\sqrt 2 \cos \theta - 1 = 0$
is

$4\cos^2\theta-2\sqrt{2}\cos \theta-1=0$

$\Rightarrow \cos\theta=\dfrac{2\sqrt{2}\pm \sqrt{(2\sqrt{2})^2+4(4)}}{2(8)}$

$=\dfrac{2\sqrt{2}\pm \sqrt{8+16}}{8}$

$\cos\theta=\dfrac{2\sqrt{2}\pm 2\sqrt{6}}{8}$

$=\dfrac{2\sqrt{2}[1\pm \sqrt{3}]}{(2)(2)(2)}=\dfrac{1+\pm \sqrt{3}}{2\sqrt{2}}$

$\cos\theta=\dfrac{1+\sqrt{3}}{2\sqrt{2}}$ or

$\cos\theta=\dfrac{1-\sqrt{3}}{2\sqrt{2}}$

$\Rightarrow \theta=\pi/12$

$\sin\left(\pi/2\right)=\dfrac{\sqrt{3}-1}{2\sqrt{2}}$

$(2\pi-\pi/2)$

$\cos\left(\pi/2-\pi/2\right)=\dfrac{1-\sqrt{3}}{2\sqrt{2}}$

$\Rightarrow \theta=\dfrac{5\pi}{12}, \left(2\pi-\dfrac{5\pi}{12}\right)$

$\Rightarrow \theta=\pi/12, \dfrac{23\pi}{12}$

$\Rightarrow \theta=\left(\dfrac{5\pi}{12}, \dfrac{19\pi}{12}\right)$

$\therefore \theta=\left\{ \dfrac{\pi}{12}, \dfrac{5\pi}{12} , \dfrac{19\pi}{12}, \dfrac{23\pi}{12}\right\}$

Solve for $x$ and $y$ :
${ x }^{ 2 }+ 2x \sin { xy } + 1= 0$

${ x }^{ 2 }+2xsin\left( xy \right) +1=0\quad \longrightarrow \left( 1 \right)$

We know that,

${ \sin }^{ 2 }\theta +{ \cos }^{ 2 }\theta =1\quad \longrightarrow \left( 2 \right)$

Substituting

$(2)$ in

$(1)$ we get

$\Rightarrow { x }^{ 2 }+2xsinxy+{ \sin }^{ 2 }xy+{ \cos }^{ 2 }xy=0$

$\Rightarrow { x }^{ 2 }+2x\sin\left( xy \right) +{ \sin }^{ 2 }\left( xy \right) +1-{ \sin }^{ 2 }\left( xy \right) =0$

$\Rightarrow \left( { x }^{ 2 }+2x\sin\left( xy \right) +{ \sin }^{ 2 }\left( xy \right) \right) =\left( { \sin }^{ 2 }\left( xy \right) -1 \right)$

$\Rightarrow { \left( x+sin\left( xy \right) \right) }^{ 2 }=\left( { sin }^{ 2 }\left( xy \right) -1 \right)$

We see that the left hand side of the above equality is always greater than zero since the square of a quantity is always greater than equal to zero. Whereas the right hand side of the quantity is always than equal to zero.

$\because$

$0\le { sin }^{ 2 }\left( xy \right) \le 1$

$\Rightarrow -1\le { \sin }^{ 2 }\left( xy \right) -1\le 0$

For the quantity to hold true we need to have

$x+sin(xy)=0$

$\longrightarrow \left( 3 \right)$ and

${ sin }^{ 2 }\left( xy \right) =1$

$\Rightarrow$

$sin\left( xy \right) =\pm 1\quad \longrightarrow \left( 4 \right)$

From

$(4)$ we have

$xy=\pi /2$ or

$-\pi /2$

For,

$xy=\pi /2$

Substituting in

$(3)$ we get

$x+1=0$

$\Rightarrow x=-1$

$\therefore$

$y=-\pi /2$

Similarly for

$xy=-\pi /2$

Substituting in

$(3)$ we get,

$x-1=0$

$\Rightarrow x=1$

and

$y=\pi /2$

$\therefore$ the possible values are

$x=-1$ and

$y=-\pi /2$ and

$x=1$ and

$y=\pi /2$

If $\sum_{i=1}^{n}cos \Theta _{i}=n,$ then the value of $\sum_{i=1}^{n}sin\Theta _{i}$

Since

$\sum_{i=1}^{n}cos \theta _{i}=n$

$\rightarrow$

$Cos\theta_{i}=1 (i=1,2,....,n)$

$\rightarrow$

$Sin\theta_{i}=0 (i=1,2,....,n)$

$\rightarrow$

$\sum_{i=1}^{n}sin \theta _{i}=0\times n=0$

Find the value of sec $\theta$ -tan $\theta$ if
sec $\theta$ +tan $\theta$ =5 ,

We know that sec

$^{2}\theta$ -tan

$^{2}\theta$ =1

$\rightarrow$

$(sec\theta-tan\theta)(5)=1$

So,

$Sec\theta-tan\theta=\dfrac{1}{5}$

$\cos 2 \hat { x } + a \sin x = 2 a - 7$ has a solution for what values of a?

$\\cos2x+asinx=2a-7\\1-2sin^2x+asinx=2a-7\\2sin^2x-asinx+(2a-8)=0\\for\>it\>to\>have\>real\>solution\>D\geq\>0\\(-a)^2-4\times2\times(2a-8)\geq\>0\\a^2-16a+64\geq\>0\\(a-8)^2\geq\>0\\which\>is\>valid\>for\>any\>value\>of\>a\\now\\-1\leqslant\>sinx\>\leqslant\>1\\-1\leqslant\>(\frac{-(-a)\pm\>\sqrt{D}}{2\times2})\leqslant\>1\\-4\>\leqslant\>a\pm\>|a-8|\leqslant\>4\\if\>a>8\\then\\-4\leqslant\>a+a-8\leqslant\>4\\-4\>\leqslant\>2a-8\leqslant\>4\\4\leqslant\>2a\leqslant\>12\\2\>\leqslant\>a\>\leqslant\>6(not\>possible)\\if\>a<8\\then\\-4\leqslant\>a-a+8\leqslant\>4\\-4\leqslant\>8\leqslant\>4(not\>possible)\\hence\>for\>no\>vlaue\>of\>a,\>equation\\will\>have\>solution$

Show that $\dfrac { 1 }{ \text{cosec}\theta -\cot\theta } -\dfrac { 1 }{ \sin\theta } =\dfrac { 1 }{ \sin\theta } -\dfrac { 1 }{ \text{cosec}\theta +\cot\theta }$

We know,

$\csc^{2}\theta-\cot^{2}\theta=1$

$\therefore \cfrac{1}{\csc\theta-\cot\theta}=\csc\theta+\cot\theta$

$\therefore \cfrac{1}{\csc\theta-\cot\theta}-\cfrac{1}{\sin\theta}=\csc\theta+\cot\theta-\csc\theta=\cot\theta$

Now,

$\cfrac{1}{\csc\theta+\cot\theta}=\csc\theta-\cot\theta$

$\therefore \cfrac{1}{\sin\theta}-\cfrac{1}{\csc\theta+\cot\theta}=\csc\theta-(\csc\theta-\cot\theta)=\cot\theta$

$\therefore \cfrac{1}{\csc\theta-\cot\theta}-\cfrac{1}{\sin\theta}=\cfrac{1}{\sin\theta}-\cfrac{1}{\csc\theta+\cot\theta}$

Prove the given statement :
$\cos^{4} A - \sin^{4} A + 1 = 2\cos^{2} A$

$cos^{4} A-sin^{4} A +1$

$=(\sin ^2A+\cos ^2A)(\cos ^2A-\sin ^2 A)+1\\ =\cos ^2A-\sin^2A+1\\ =\cos ^2A+\cos ^2A\\=2\cos ^2A$

Simplify:-
${\left( {\text{cos ec}\theta - \cot \theta } \right)^2}$

1208406_1235528_ans_eda4b36026084b99bfcaefda186001af.jpg

Prove:
$\cot ^{ 2 }{ A\left( \dfrac { \sec { A-1 } }{ 1+\sin { A } } \right) } +\sec ^{ 2 }{ A\left( \dfrac { \sin { A-1 } }{ 1+\sec { A } } \right) } =0$

${ cot }^{ 2 }A\left( \dfrac { secA-1 }{ 1+sinA } \right) +{ sec }^{ 2 }A\left( \dfrac { sinA-1 }{ 1+secA } \right)$

Rationalising the above L.H.S,

$\Longrightarrow { cot }^{ 2 }A\left( \dfrac { (secA-1)(1-sinA) }{ (1+sinA)(1-sinA) } \right) +{ sec }^{ 2 }A\left( \dfrac { (sinA-1)(1-secA) }{ (1+secA)(1-secA) } \right)$

$\Longrightarrow { cot }^{ 2 }A\left( \dfrac { (secA-secAsinA-1+sinA) }{ (1-{ sin }^{ 2 }A) } \right) +{ sec }^{ 2 }A\left( \dfrac { (secA-secAsinA-1+sinA) }{ (1-{ sec }^{ 2 }A) } \right)$

$\Longrightarrow \dfrac { { cos }^{ 2 }A }{ { sin }^{ 2 }A } \left( \dfrac { (secA-secAsinA-1+sinA) }{ ({ cos }^{ 2 }A) } \right) +\dfrac { 1 }{ { cos }^{ 2 }A } \left( \dfrac { (secA-secAsinA-1+sinA) }{ ({ tan }^{ 2 }A) } \right)$

Since,

$1-{ sin }^{ 2 }A={ cos }^{ 2 }A$

$\Longrightarrow \dfrac { (secA-secAsinA-1+sinA) }{ ({ sin }^{ 2 }A) } +\dfrac { (secA-secAsinA-1+sinA) }{ ({ sin }^{ 2 }A) }$

$\Longrightarrow \dfrac { (secA-secAsinA-1+sinA+secA-secAsinA-1+sinA) }{ ({ sin }^{ 2 }A) }$

$\Longrightarrow \dfrac { 0 }{ { sin }^{ 2 }A }$

$\Longrightarrow 0$

$\Longrightarrow R.H.S$

Find the principle solution of the equation .
$\sin x=-\dfrac{1}{2}$

We have,

$\sin x=\dfrac{-1}{2}$

$\Rightarrow \sin x= \sin \dfrac{\pi}{6}$

$\Rightarrow \sin x = \sin \left( \pi+\dfrac{\pi}{6} \right) \because \sin (\pi+ \theta)=-\sin \theta$

$\Rightarrow \sin x= \sin \dfrac{\pi}{6}$

Then,

$x=7\dfrac{\pi}{6}$

Again,

$\sin x=\dfrac{-1}{2}$

$\sin x=-5 \sin \dfrac{\pi}{6}$

$\sin x= \sin \left( 2\pi-\dfrac{\pi}{6} \right)……. \sin (360^{o}-\theta)=\sin \theta$

$\sin x= \sin 11 \dfrac{\pi}{6}$

Then,

$x=11 \dfrac{\pi}{6}$

Hence, this is the principle value of

$\sin x= -\dfrac{1}{2}$

Prove that:
$2 + \tan ^ { 2 } \theta + \dfrac { 1 } { \tan ^ { 2 } \theta } = \dfrac { 1 } { \sin ^ { 2 } \theta - \sin ^ { 4 } \theta }$

$2+tan^{2}\theta +\dfrac{1}{tan^{2}\theta }$

$\Rightarrow 2+\dfrac{\sin^{2}\theta }{\cos^{2}\theta }+\dfrac{1}{\dfrac{\sin^{2}\theta }{\cos^{2}\theta }}$

$\Rightarrow \dfrac{2(\cos^{2}\theta )+\sin^{2}\theta }{\cos^{2}\theta }+\dfrac{\cos^{2}\theta }{\sin^{2}\theta }$

$\Rightarrow \dfrac{2(1-\sin^{2}\theta )+\sin^{2}\theta }{(1-\sin^{2}\theta )}+\dfrac{(1-\sin^{2}\theta )}{\sin^{2}\theta }$

$\Rightarrow \dfrac{2-2\sin^{2}\theta +\sin^{2}\theta }{1-\sin^{2}\theta }+\dfrac{1-\sin^{2}\theta }{\sin^{2}\theta }$

$\Rightarrow \dfrac{2\sin^{2}\theta -\sin^{4}\theta +1+\sin^{4}\theta -2\sin^{2}\theta }{\sin^{2}\theta -\sin^{4}\theta }$

$\Rightarrow \dfrac{1}{\sin^{2}\theta -\sin^{4}\theta }$ Hence proved.

If $m\cos^2A+n\sin^2A=p$ , then $\cot^2A=$

$\quad \quad m\cos ^{ 2 }{ A } +n\sin ^{ 2 }{ A } =p\\ \Rightarrow m\left( 1-\sin ^{ 2 }{ A } \right) +n\sin ^{ 2 }{ A } =p\\ \Rightarrow m+\left( n-m \right) \sin ^{ 2 }{ A } =p\\ \Rightarrow \sin ^{ 2 }{ A } =\frac { p-m }{ n-m } \quad \quad \quad .......(1)\\ \Rightarrow 1-\cos ^{ 2 }{ A } =\frac { p-m }{ n-m } \\ \Rightarrow \cos ^{ 2 }{ A } =1-\frac { p-m }{ n-m } \\ \Rightarrow \cos ^{ 2 }{ A } =\frac { n-m-p+m }{ n-m } \\ \Rightarrow \cos ^{ 2 }{ A } =\frac { n-p }{ n-m } \quad \quad \quad .....(2)\\$

from

$(1)$ and

$(2)$

$\cot ^{ 2 }{ A } =\frac { \cos ^{ 2 }{ A } }{ \sin ^{ 2 }{ A } } \\ \Rightarrow \cot ^{ 2 }{ A } =\frac { \frac { n-p }{ n-m } }{ \frac { p-m }{ n-m } } \\ \Rightarrow \cot ^{ 2 }{ A } =\frac { n-p }{ p-m }$

$\dfrac { { cos }^{ 2 }{ 20 }^{ \circ }+{ cos }^{ 2 }{ 70 }^{ \circ } }{ { sin }^{ 2 }{ 31 }^{ \circ + }+{ sin }^{ 2 }{ 59 }^{ \circ } } +{ sin }^{ 2 }{ 64 }^{ \circ }+cos{ 64 }^{ \circ }sin{ 26 }^{ \circ }$

$\cfrac { \cos ^{ 2 }{ { 20 }^{ o } } +\cos ^{ 2 }{ { 70 }^{ o } } }{ \sin ^{ 2 }{ 31^o } +\sin ^{ 2 }{ 59^o } } +\sin ^{ 2 }{ { 64 }^{ o } } +\cos ^{ 2 }{ { 64 }^{ o } } \sin ^{ 2 }{ { 26 }^{ o } }$

$\cfrac { \sin ^{ 2 }{ { 70 }^{ o } } +\cos ^{ 2 }{ { 70 }^{ o } } }{ \sin ^{ 2 }{ 31^o } +\cos ^{ 2 }{ 31^o } } +\sin ^{ 2 }{ { 64 }^{ o } } +\cos ^{ 2 }{ { 64 }^{ o } }$

$1+1=2$

Prove the identity: $\dfrac{{{\text{sinA}}}}{{{\text{1 + cosA}}}} = {\text{coscA}} - {\text{cotA}}$ .

LHS

$=\cfrac{\sin A}{1+\cos A}$

$=\cfrac{\sin A(1-\cos A)}{(1+\cos A)(1-\cos A)}$

$=\cfrac{\sin A-\sin A\cos A}{1-\cos^2 A}$

$=\cfrac{`1}{\sin A}-\cfrac{\cos A}{\sin A}$

$=\csc A-\cot A=$ RHS

If $'\theta'$ is the parameter, then the family of lines represented by
$(2 cos \theta + 3 sin\theta) X + (3 cos \theta - 5 sin \theta)$
y - $(5cos\theta - 7 sin\theta)$ = 0 are concurrent at the point

$cos\theta (2x+3y-5)+sin\theta (3x-5y+7) = 0$

$\Rightarrow (2x+3y-5)+tan\theta (3x-5y+7) = 0$

of from

$L_{1}+\lambda L_{2} = 0,$ intersection point be P

$3\times (2x+3y-5=0)$

$2\times (3x-5y+7=0)$

___________________

$19y-29 = 0\Rightarrow y = \frac{29}{19}$

$\Rightarrow x=\frac{5y-7}{3}=\frac{145-133}{57}=\frac{12}{57}$

$\Rightarrow P(-\frac{4}{19},\frac{29}{19})$

1167328_1248431_ans_305650ccc414485992aa20f10e33f680.jpg

Prove that $\cot \theta - \tan \theta = \dfrac{{2{{\cos }^2}\theta - 1}}{{\sin \theta \cos \theta }}$

RHS

$=\dfrac{2\cos^2\theta -1}{\sin\theta \cos\theta}=\dfrac{\cos^2\theta +\cos^2\theta -1}{\sin\theta \cos\theta}$

$=\dfrac{\cos^2\theta -(1-\cos^2\theta)}{\sin\theta \cos \theta}$

$=\dfrac{\cos^2\theta -\sin^2\theta}{\sin\theta \cos\theta}$

$=\dfrac{\cos\theta}{\sin\theta}-\dfrac{\sin\theta}{\cos\theta}$

$=\cot\theta -\tan\theta$

$=$ LHS.

Prove that: $\dfrac { \tan\theta +\sec\theta -1 }{ \tan\theta -\sec\theta -1 } =\sec\theta +\tan\theta \\$

LHS

$=\dfrac { \tan\theta +\sec\theta -1 }{ \tan\theta -\sec\theta -1 }$

$=\dfrac { \tan\theta +\sec\theta -{ \sec }^{ 2 }\theta +{ \tan }^{ 2 }\theta }{ \tan\theta -\sec\theta }$

$=\dfrac { \left( \tan\theta +\sec\theta \right) -\left( \sec\theta +\tan\theta \right) \left( \sec\theta -\tan\theta \right) }{ \left( \tan\theta -\sec\theta \right) }$

$=\dfrac { \left( \tan\theta +\sec\theta \right) \left( 1-\sec\theta +\tan\theta \right) }{ \left( \tan\theta -\sec\theta +1 \right) }$

$=\tan\theta +\sec\theta$ (RHS)

(proved)

Prove : $\dfrac {\cos^{3}\theta-\sin^{3}\theta}{\cos \theta-\sin \theta}+\dfrac {\cos^{3}\theta+\sin^{3}\theta}{\cos \theta+\sin \theta}=2$

Proof :

$LHS : \dfrac { \cos ^{ 3 }{ \theta } -\sin ^{ 3 }{ \theta } }{ \cos { \theta } -\sin { \theta } } +\dfrac { \cos ^{ 3 }{ \theta } +\sin ^{ 3 }{ \theta } }{ \cos { \theta } +\sin { \theta } }$

we use two identities,

$\Rightarrow a^3-b^3=\left( a-b\right) \left( a^2+b^2+ab\right)$

$\Rightarrow a^3+b^3=\left( a+b\right) \left( a^2+b^2-ab\right)$

using the above identities,

$=\dfrac { \left( \cos { \theta } -\sin { \theta } \right) \left( \cos ^{ 2 }{ \theta } -\sin ^{ 2 }{ \theta } +\cos { \theta } .\sin { \theta } \right) }{ \left( \cos { \theta } -\sin { \theta } \right) }+\dfrac { \left( \cos { \theta }+\sin { \theta } \right) \left( \cos ^{ 2 }{ \theta } -\sin ^{ 2 }{ \theta } -\cos { \theta } .\sin { \theta } \right) }{ \left( \cos { \theta } +\sin { \theta } \right) }$

$=\cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } +\cos { \theta } .\sin { \theta } +\cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } -\cos { \theta } .\sin { \theta }$

$=2\left( \cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } \right)$

$=2$

$\left( \because \cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } =1\right)$

Hence, proved.

Prove that $\cot \theta - \tan \theta = \dfrac{{\left( {2{{\cos }^2}\theta - 1} \right)}}{{\left( {\sin \theta. \cos\theta } \right)}}$

$\begin{array}{l} Use\, \, the\, \, identities, \\ 2{ \cos ^{ 2 } }\theta -1=\cos 2\theta ={ \cos ^{ 2 } }\theta -{ \sin ^{ 2 } }\theta . \\ Now, \\ LHS=\cot \theta -\tan \theta =\dfrac { { \cos \theta } }{ { \sin \theta } } -\dfrac { { \sin \theta } }{ { \cos \theta } } \\ LHS=\cot \theta -\tan \theta =\dfrac { { { { \cos }^{ 2 } }\theta -{ { \sin }^{ 2 } }\theta } }{ { \sin \theta \cos \theta } } \\ LHS=\cot \theta -\tan \theta =\dfrac { { 2{ { \cos }^{ 2 } }\theta -1 } }{ { \sin \theta \cos \theta } } =RHS \end{array}$

Hence, proved.

If $cos\left( \alpha \right) =\dfrac { 1 }{ 2 }$ then $\alpha=?$

1988257_1268688_ans_e4b5de19a64d4057a6a857459a30f041.jpg

Given that $tan( \theta_1+ \theta_2)= \dfrac{tan \theta_1+ tan \theta_2}{1- tan \theta_1tan \theta_2}$ . Find $( \theta_1+ \theta_2)$ when $tan \theta_1= \dfrac 12$ $tan \theta_2=\dfrac 13$ .

1338009_1274636_ans_052d310a663140bd85c88b2dca3de7e6.PNG

Evaluate: $\tan ^ { 2 } \dfrac { \pi } { 16 } + \tan ^ { 2 } \dfrac { 2 \pi } { 16 } + \tan ^ { 2 } \dfrac { 3 \pi } { 16 } + \ldots \ldots + \tan ^ { 2 } \dfrac { 7 \pi } { 16 }$

$\dfrac{7\pi}{16}=\dfrac{\pi}{2}-\dfrac{\pi}{16}$

$\dfrac{6\pi}{16}=\dfrac{\pi}{2}-\dfrac{2\pi}{16}$

$\dfrac{5\pi}{16}=\dfrac{\pi}{2}-\dfrac{2\pi}{16}$

$\dfrac{4\pi}{16}=\dfrac{\pi}{4}$

$\Rightarrow \tan^2\dfrac{\pi}{16}+\tan^2\dfrac{2\pi}{16}+\tan^2\dfrac{3\pi}{16}+...…..\tan^2\dfrac{6\pi}{16}+\tan^2\dfrac{7\pi}{16}$

$\Rightarrow \tan^2\dfrac{\pi}{16}+\tan^2\dfrac{2\pi}{16}+\tan^2\dfrac{3\pi}{16}+...…+\tan^2\left(\dfrac{\pi}{3}-\dfrac{2\pi}{16}\right)+\tan^2\left(\dfrac{\pi}{2}-\dfrac{\pi}{16}\right)$

$\Rightarrow \tan^2\dfrac{\pi}{16}+\tan^2\dfrac{2\pi}{16}\tan^2\dfrac{3\pi}{16}+...…..+\cot^2\dfrac{2\pi}{16}+\cot^2\dfrac{\pi}{16}$

$\Rightarrow \left(\tan^2\dfrac{\pi}{16}+\cot^2\dfrac{\pi}{16}\right)+\left(\tan^2\dfrac{2\pi}{16}+\cot^2\dfrac{2\pi}{16}\right)+\left(\tan^2\dfrac{3\pi}{16}+\cot^2\dfrac{3\pi}{16}\right)+\tan^2\dfrac{4\pi}{16}$

Solve first bracket

$\left(\tan^2\dfrac{\pi}{16}+\cot^2\dfrac{\pi}{16}\right)=\left(\tan\dfrac{\pi}{16}+\cot\dfrac{\pi}{16}\right)^2-2\tan\dfrac{\pi}{16}\cot\dfrac{\pi}{16}$

$=\left(\dfrac{\sin \pi/16}{\cos \pi/16}+\dfrac{\cos \pi/16}{\sin \pi/16}\right)^2-2$

$=\left(\dfrac{t\times 2}{2\sin(\pi/16)\cos(\pi/16)}\right)^2-2$

$=\left(\dfrac{2}{\sin\pi/8}\right)^2-2$

$=\dfrac{4\times 2}{2\sin^2\pi/8}-2$

$=\dfrac{8}{(1-\cos \pi/4)}-2$

$=\dfrac{8}{\left(1-\dfrac{1}{\sqrt{2}}\right)}-2$

$=\dfrac{8\sqrt{2}}{(\sqrt{2}-1)}-2$

$\Rightarrow 8\sqrt{2}(\sqrt{2}+1)-2$

Similarly

$2^{nd}$ bracket

$=\dfrac{8}{\left(1-\cos\dfrac{\pi}{2}\right)}-2=8-2=6$

& similarly

$3^{rd}$ bracket

$=\dfrac{8}{(1-\cos 3\pi/4)}-2=8\sqrt{2}(\sqrt{2}-1)-2$

Thus, equation

$(1)$

$\Rightarrow 8\sqrt{2}(\sqrt{2}+1)-2+6+1+8\sqrt{2}(\sqrt{2}-1-2=$

$\Rightarrow 16+8\sqrt{2}-2+7+8\sqrt{2}\times \sqrt{2}-8\sqrt{2}-2=$

$\Rightarrow 35$ .

1188977_1291735_ans_74f1216f6a394fd184b7d4edda77ec64.jpg

If $cosA=\dfrac{\sqrt{3}}{2}$ , find the value of cotA+ tanA

2063175_1279289_ans_c29b44b8d5854973a75d33f9a871fa9e.jpg

Prove that $\sin^6\theta + {\cos ^6}\theta + 3\sin^2\theta \cos^2\theta = 1$

We know that

$\sin^2\theta +\cos^2\theta =1$

Cubing both sides

$(\sin^2\theta +\cos^2\theta)^3=1^3$

$(\sin^2\theta)^3+(\cos^2\theta)^3+3\sin^2\theta\cos^2\theta(\sin^2\theta +\cos^2\theta)=1$

$\sin^6\theta +\cos^6\theta +3\sin^2\theta \cos^2\theta (1)=1$

$\sin^6\theta +\cos^6\theta +3\sin^2\theta \cos^2\theta =1$

Hence proved.

Solve
$\dfrac {\sin^{2}\theta}{\cos^{2}\theta}-\dfrac {\sin^{2}\theta}{1}=\tan^{2}\theta \cdot \sin^{2}\theta$

$\frac{sin^{2}\theta }{cos^{2}\theta }= sin^{2}\theta = tan^{2}\theta sin^{2}\theta$

$sin^{2}\theta (\frac{1}{cos^{2}\theta }-1)=\frac{sin^{2}\theta \times sin^{2}\theta }{cos^{2}\theta }$

such that

$sin^{2}\theta \neq 0$ &

$cos^{2}\theta \neq 0$

we get ,

$1-cos^{2}\theta = sin^{2}\theta$ Hence proved.

1169193_1307209_ans_8dff12954c434e7eb6d8f479fca22d85.jpg

If $\sin\left( x-20 \right) ^{ \circ }\cos\left( 3x-10 \right) ^{ \circ }=0$ , Then find the value of x.

We have,

$\sin \left( x-{{20}^{o}} \right)\cos \left( 3x-{{10}^{o}} \right)=0$

Then,

$\sin \left( x-{{20}^{o}} \right)\cos \left( x-{{30}^{o}} \right)=0$

$\Rightarrow \sin \left( x-{{20}^{o}} \right)=0,\,\,\cos \left( x-{{20}^{o}} \right)=0$

$\Rightarrow \sin \left( x-{{20}^{o}} \right)=\sin 0,\,\,\cos \left( x-{{20}^{o}} \right)=\cos \dfrac{\pi }{2}$

$\Rightarrow x-{{20}^{o}}=0,\,\,\,x-{{20}^{o}}=\dfrac{\pi }{2}$

$\Rightarrow x={{20}^{o}},\,\,\,x-{{20}^{o}}={{90}^{o}}$

$\Rightarrow x={{20}^{o}},\,\,\,\,x={{70}^{o}}$

Hence, this is the answer.

If $A-B=\dfrac{\pi}{4}$ , then $(1+\tan{A})(1-\tan{B})=$

$(1+\tan A)(1-\tan B)= 1+\tan A-\tan B-\tan A\tan B$

we know

$\tan (A-B)=\dfrac {\tan A-\tan B}{1+\tan A\tan B}$

$1=\dfrac {tanA-tanB}{1+tanAtanB}$

$\tan A-\tan B=1+\tan A\tan B$

$(1+\tan A)(1-\tan B)= 1+\tan A-\tan B-\tan A\tan B$

$=1+1+\tan A\tan B-\tan A\tan B$

$=2$

If $\cot \theta =\dfrac {7}{8}$ , evaluate:
$\dfrac {(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$ .

$\begin{array}{l} \cot \theta =\dfrac { 7 }{ 8 } \\ \dfrac { { \left( { 1+sin\theta } \right) \left( { 1-sin\theta } \right) } }{ { \left( { 1+\cos \theta } \right) \left( { 1-\cos \theta } \right) } } \\ =\dfrac { { { 1^{ 2 } }-{ sin^{ 2 } }\theta } }{ { { 1^{ 2 } }-{ { \cos }^{ 2 } }\theta } } =\dfrac { { { { \cos }^{ 2 } }\theta } }{ { { { \sin }^{ 2 } }\theta } } \\ ={ \cot ^{ 2 } }\theta ={ \left( { \dfrac { 7 }{ 8 } } \right) ^{ 2 } }=\dfrac { { 49 } }{ { 64 } } \end{array}$

If ${\tan ^{ - 1}}\sqrt 3 + {\cot ^{ - 1}}x = \frac{\pi }{2}$ then find the value of $x$ .

We know that

${\tan}^{-1}{\sqrt{3}}=\dfrac{\pi}{3}$

$\therefore\,{\tan}^{-1}{\sqrt{3}}+{\cot}^{-1}{x}=\dfrac{\pi}{2}$

$\Rightarrow\,\dfrac{\pi}{3}+{\cot}^{-1}{x}=\dfrac{\pi}{2}$

$\Rightarrow\,{\cot}^{-1}{x}=\dfrac{\pi}{2}-\dfrac{\pi}{3}=\dfrac{3\pi-2\pi}{6}=\dfrac{\pi}{6}$

$\Rightarrow\,x=\cot{\dfrac{\pi}{6}}=\sqrt{3}$

$\therefore\,x=\sqrt{3}$

Solve:
$(\sin \theta+\cos \theta)(\tan \theta+\cot \theta)=\csc \theta+\cot \theta$

We have,

$(\sin \theta+\cos \theta)(\tan \theta+\cot \theta)=\csc \theta+\cot \theta$

$\Rightarrow \left( \sin \theta +\cos \theta \right)\left( \dfrac{\sin \theta }{\cos \theta }+\dfrac{\cos \theta }{\sin \theta } \right)=\csc \theta +\cot \theta$

$\Rightarrow \left( \sin \theta +\cos \theta \right)\left( \dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{\sin \theta \cos \theta } \right)=\csc \theta +\cot \theta$

$\Rightarrow \left( \sin \theta +\cos \theta \right)\left( \dfrac{1}{\sin \theta \cos \theta } \right)=\csc \theta +\cot \theta$

$\Rightarrow \dfrac{\sin \theta +\cos \theta }{\sin \theta \cos \theta }=\csc \theta +\cot \theta$

$\Rightarrow \dfrac{\sin \theta }{\sin \theta \cos \theta }+\dfrac{\cos \theta }{\sin \theta \cos \theta }=\csc \theta +\cot \theta$

$\Rightarrow \sec \theta +\csc \theta =\csc \theta +\cot \theta$

$\Rightarrow \sec \theta =\cot \theta$

$\Rightarrow \dfrac{1}{\cos \theta }=\dfrac{\cos \theta }{\sin \theta }$

$\Rightarrow {{\cos }^{2}}\theta =\sin \theta$

$\Rightarrow 1-{{\sin }^{2}}\theta ={{\sin }^{2}}\theta$

$\Rightarrow 1=2{{\sin }^{2}}\theta$

$\Rightarrow \dfrac{1}{2}={{\sin }^{2}}\theta$

$\Rightarrow {{\sin }^{2}}\theta =\dfrac{1}{2}$

$\Rightarrow {{\sin }^{2}}\theta ={{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}$

On comparing both side and we get,

${{\sin }^{2}}\theta ={{\sin }^{2}}\dfrac{\pi }{4}$

$\theta =\dfrac{\pi }{4}\,\,\left( \text{Principal}\,\text{value} \right)$

Hence, this is the answer.

Prove that $\displaystyle \frac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \frac{1}{{\sec \theta - \tan \theta }}$ , using the identity ${\sec ^2} = 1 + {\tan ^2}\theta .$

1213376_1305932_ans_5c9eb00c76b34e8d9eb356a2f857eb56.JPG

Prove that, ${ cos20 }^{ \circ }{ cos }40^{ \circ }{ cos60 }^{ \circ }{cos80^o}=\dfrac { 1 }{ 16 }$

We have,

$L.H.S.$

$\cos {{20}^{o}}\cos {{40}^{o}}\cos {{60}^{o}}\cos {{80}^{o}}$

$\Rightarrow \cos {{20}^{o}}\cos {{40}^{o}}\dfrac{1}{2}\cos {{80}^{o}}\,\,\,\,\,\therefore \cos {{60}^{o}}=\dfrac{1}{2}$

$\Rightarrow \dfrac{1}{2}\cos {{20}^{o}}\cos {{40}^{o}}\cos {{80}^{o}}$

Multiply and dived by 2 and we get,

$\Rightarrow \dfrac{1}{4}\left( 2\cos {{20}^{o}}\cos {{40}^{o}} \right)\cos {{80}^{o}}$

$\Rightarrow \dfrac{1}{4}\left( \cos \left( {{20}^{o}}+{{40}^{o}} \right)+\cos \left( {{20}^{o}}-{{40}^{o}} \right) \right)\cos {{80}^{o}}$

$\Rightarrow \dfrac{1}{4}\left( \cos {{60}^{o}}+\cos {{20}^{o}} \right)\cos {{80}^{o}}$

$\Rightarrow \dfrac{1}{4}\left( \dfrac{1}{2}+\cos {{20}^{o}} \right)\cos {{80}^{o}}$

$\Rightarrow \dfrac{1}{8}\left( 1+2\cos {{20}^{o}} \right)\cos {{80}^{o}}$

$\Rightarrow \dfrac{1}{8}\left( \cos {{80}^{o}}+2\cos {{20}^{o}}\cos {{80}^{o}} \right)$

$\Rightarrow \dfrac{1}{8}\left( \cos {{80}^{o}}+\left( \cos \left( {{20}^{o}}+{{80}^{o}} \right)+\cos \left( {{20}^{o}}-{{80}^{o}} \right) \right) \right)$

$\Rightarrow \dfrac{1}{8}\left( \cos {{80}^{o}}+\left( \cos {{100}^{o}}+\cos \left( -{{60}^{o}} \right) \right) \right)$

$\Rightarrow \dfrac{1}{8}\left( \cos {{80}^{o}}+\left( \cos {{100}^{o}}+\cos {{60}^{o}} \right) \right)$

$\Rightarrow \dfrac{1}{8}\left( \cos {{80}^{o}}+\left( \cos {{100}^{o}}+\dfrac{1}{2} \right) \right)$

$\Rightarrow \dfrac{1}{8}\left( \cos {{80}^{o}}+\cos {{100}^{o}}+\dfrac{1}{2} \right)$

$\Rightarrow \dfrac{1}{8}\left( \cos {{80}^{o}}+\cos {{100}^{o}} \right)+\dfrac{1}{16}$

$\Rightarrow \dfrac{1}{8}\left( 2\cos \left( \dfrac{{{80}^{o}}+{{100}^{o}}}{2} \right)\cos \left( \dfrac{{{80}^{o}}-{{100}^{o}}}{2} \right) \right)+\dfrac{1}{16}$

$\Rightarrow \dfrac{1}{8}\left( 2\cos {{90}^{o}}\cos {{20}^{o}} \right)+\dfrac{1}{16}$

$\Rightarrow \dfrac{1}{8}\left( 2\times 0\cos {{20}^{o}} \right)+\dfrac{1}{16}$

$\Rightarrow \dfrac{1}{16}$

R.H.S.

Prove that, ${ sin20 }^{ \circ }{ sin }40^{ \circ }{ sin80 }^{ \circ }=\dfrac { \sqrt { 3 } }{ 8 }$

We have,

L.H.S.

$\sin {{20}^{o}}\sin {{40}^{o}}\sin {{80}^{o}}$

On multiplying and divide by 2 and we get,

$\Rightarrow \dfrac{1}{2}\left[ 2\sin {{20}^{o}}\sin {{40}^{o}}\sin {{80}^{o}} \right]$

$\Rightarrow \dfrac{1}{2}\left[ \cos \left( {{20}^{o}}-{{40}^{o}} \right)-\cos \left( {{20}^{o}}+{{40}^{o}} \right) \right]\sin {{80}^{o}}$

$\Rightarrow \dfrac{1}{2}\left[ \cos {{20}^{o}}-\cos {{60}^{o}} \right]\sin {{80}^{o}}$

$\Rightarrow \dfrac{1}{2}\left[ \cos {{20}^{o}}-\dfrac{1}{2} \right]\sin {{80}^{o}}$

$\Rightarrow \dfrac{1}{4}\left[ 2\cos {{20}^{o}}-1 \right]\sin {{80}^{o}}$

$\Rightarrow \dfrac{1}{4}\left[ 2\cos {{20}^{o}}\sin {{80}^{o}}-\sin {{80}^{o}} \right]$

$\Rightarrow \dfrac{1}{4}\left[ \sin \left( {{20}^{o}}+{{80}^{o}} \right)-\sin \left( {{20}^{o}}-{{80}^{o}} \right)-\sin {{80}^{o}} \right]$

$\Rightarrow \dfrac{1}{4}\left[ \sin {{100}^{o}}+\sin {{60}^{o}}-\sin {{80}^{o}} \right]$

$\Rightarrow \dfrac{1}{4}\left[ \sin {{100}^{o}}-\sin {{80}^{o}}+\dfrac{\sqrt{3}}{2} \right]$

$\Rightarrow \dfrac{1}{4}\left[ 2\cos \left( \dfrac{{{100}^{o}}+{{80}^{o}}}{2} \right)\sin \left( \dfrac{{{100}^{o}}-{{80}^{o}}}{2} \right)+\dfrac{\sqrt{3}}{2} \right]$

$\Rightarrow \dfrac{1}{4}\left[ 2\cos {{90}^{o}}\sin {{10}^{o}}+\dfrac{\sqrt{3}}{2} \right]$

$\Rightarrow \dfrac{1}{4}\left[ 2\times 0\times \sin {{10}^{o}}+\dfrac{\sqrt{3}}{2} \right]$

$\Rightarrow \dfrac{\sqrt{3}}{8}$

R.H.S.

Prove ;
$\dfrac{\tan A}{(1+\tan^{2}A)^{2}}+\dfrac{\cot{A}}{(1+\cot^{2}A)^{2}}=\sin A \cos A$

Taking L.H.S.-

$\cfrac{\tan{A}}{{\left( 1 + \tan^{2}{A} \right)}^{2}} + \cfrac{\cot{A}}{{\left( 1 + \cot^{2}{A} \right)}^{2}}$

$= \cfrac{\tan{A}}{{\left( \sec^{2}{A} \right)}^{2}} + \cfrac{\cot{A}}{{\left( \csc^{2}{A} \right)}^{2}}$

$= \cfrac{\left( \cfrac{\sin{A}}{\cos{A}} \right)}{\left( \cfrac{1}{\cos^{4}{A}} \right)} + \cfrac{\left( \cfrac{\cos{A}}{\sin{A}} \right)}{\left( \cfrac{1}{\sin^{4}{A}} \right)}$

$= \cos^{3}{A} \sin{A} + \sin^{3}{A} \cos{A}$

$= \sin{A} \cos{A} \left( \cos^{2}{A} + \sin^{2}{A} \right)$

$= \sin{A} \cos{A}$

$\dfrac{{\cos 5x + \cos 4x}}{{1 - 2\cos 3x}}$ $=$ $- \left( {\cos 2x + \cos x} \right)$

$LHS=\dfrac{(cos5x +cos4x)}{(1–2cos3x)}$

$=\dfrac{2cos(9x/2) cos(x/2)}{(1–2cos3x)}$

Above and below multiple by $cos(3x/2)$

$=\dfrac{(2cos9x/2 cosx/2 cos3x/2 ) }{(cos3x/2 -2cos3x cos3x/2 )}$

$=\dfrac{(2cos9x/2 cosx/2 cos3x/2 )}{(cos3x/2 -cos9x/2 -2cos3x/2 )}$

$=\dfrac{(2cos9x/2 cosx/2 cos3x/2 ) }{(-cos9x/2)}$

$=-2cos(3x/2) cos(x/2)$

$=-(cos2x +cosx) RHS.$

Prove that : $\dfrac { \left( 1 + \tan ^ { 2 } A \right) \cot A } { \csc ^ { 2 } A } = \tan A.$

$L.H.S=\dfrac { { \left( { 1-{ { \tan }^{ 2 } }A } \right) \cot A } }{ { \cos e{ c^{ 2 } }A } } \\ =\dfrac { { { { \sec }^{ 2 } }A\cot A } }{ { \cos e{ c^{ 2 } }A } } \, \, \, \, \, \, \left( { { { \sec }^{ 2 } }A=1+{ { \tan }^{ 2 } }A } \right) \\ =\dfrac { { \dfrac { 1 }{ { { { \cos }^{ 2 } }A } } \times \dfrac { { \cos A } }{ { \sin A } } } }{ { \dfrac { 1 }{ { { { \sin }^{ 2 } }A } } } } \\ =\dfrac { { \dfrac { 1 }{ { { { \cos A \sin A } } } } } }{ { \dfrac { 1 }{ { { { \sin }^{ 2 } }A } } } } \\ =\dfrac { { \sin A } }{ { \cos A } } =\tan A=R.H.S \\ Hence\, proved.$

If $x=a\sin\theta+b\cos\theta$ and $y=a\cos\theta-b\sin\theta$ , then prove that ${x}^{2}+{y}^{2}={a}^{2}+{b}^{2}$

$\begin{array}{l} x=a\sin \theta +b\cos \theta \\ y=a\cos \theta -b\sin \theta \\ Now, \\ { x^{ 2 } }+{ y^{ 2 } }={ a^{ 2 } }+{ b^{ 2 } }+2ab\sin \theta \cos \theta -2ab\sin \theta \cos \theta \\ { x^{ 2 } }+{ y^{ 2 } }={ a^{ 2 } }+{ b^{ 2 } } \\ hence,\, proved. \end{array}$

Evaluate the following.
$2\tan^{2} 45^{o}+\cos^{2} 30^{o}$

$2\tan^245^o+\cos^230^o$

$\Rightarrow 2(1)^2+\left(\dfrac{\sqrt{3}}{2}\right)^2$

$\Rightarrow 2+\dfrac{3}{4}=\dfrac{11}{4}$ .

1189517_1354839_ans_509e3d354ca14801ba433d9487d82b96.jpg

Evaluate the following.
$\dfrac {\sec ^{2}60^{o}-\tan ^{2}60^{o}}{\sin ^{2}30^{o}+\cos ^{2}30^{o}}$

$\dfrac{\sec^245^o-\tan^260^o}{\sin^230^o-\cos^245}=\dfrac{(\sqrt{2})^2-(\sqrt{3})^2}{\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{\sqrt{2}}\right)^2}$

$\Rightarrow \dfrac{2-3}{\dfrac{1}{4}-\dfrac{1}{2}}=\dfrac{-1}{-1/4}=4$ .

1189518_1354847_ans_ca2250bd0604417dac6427dfb78267c1.jpg

If sin (A+B) = 1 and sin (A-B) = $\frac { 1 }{ \sqrt { 3 } }$ , find the value of tan A + tan B.

$\sin(A+B)=1\implies A+B=\pi/2\implies A=\pi/2-B$

$\sin(A-B)=\dfrac{1}{\sqrt{3}} \implies \sin (\pi/2-2{B})=\dfrac{1}{\sqrt{3}}\implies \cos 2{B}=\dfrac{1}{\sqrt{3}}\implies \sin 2{B}=\sqrt{\dfrac{2}{3}}$

$\sin 2{B}=\sqrt{\dfrac{2}{3}}\implies \dfrac{2\tan B}{1+\tan^2 B}=\sqrt{\dfrac{2}{3}}$

$\implies \dfrac{1+\tan^2 B}{\tan B}=\sqrt{6}\implies \cot B+\tan B=\sqrt{6}\implies \cot (\pi/2-A)+\tan B=\sqrt{6}$

$\implies \tan A+\tan B=\sqrt{6}$

Prove that : $\left( {\cos ecA - \sin A} \right)\left( {\sec A - \cos A} \right) = \dfrac{1}{{\tan A + \cot A}}$

Give

$(cosA-sinA)(secA-CosA)=\frac{1}{tanA+cotA}$

L.H.S.(coscA sinA)(secA-cosA)

$\Rightarrow L.H.S=(\frac{1}{sinA}-sinA) (\frac{1}{cosA}-cosA)$

$\Rightarrow L.H.S = (\frac{1-sin^{2}A}{sinA})(\frac{1-cos^{2}A}{cosA})$

$\Rightarrow L.H.S=\frac{cos^{2}A.sin^{2}A}{sina.cosA}\left [ \because sin^{2}A+cos^{2}A-1 cos^{2}A-1-sin^{2}A sin^{2}A=1-cos^{2}A \right ]$

$\Rightarrow L.H.S=\frac{sinA.cosA}{1}$

$\Rightarrow L.H.S=\frac{sinA.cosA}{sin^{2}A+cos^{2}A}$

$\Rightarrow L.H.S=\frac{1}{\frac{sin^{2}A+cos^{2}A}{sinA.cosA}}$

$\Rightarrow L.H.S=\frac{1}{\frac{sin^{2}A}{sinA cosA}+\frac{cos^{2}A}{sinA.cosA}}$

$\Rightarrow L.H.S=\frac{1}{\frac{sinA}{cosA}+\frac{cosA}{sinA}}$

$\Rightarrow L.H.S=\frac{1}{tanA+cotA}$

=R.H.S

$\Rightarrow$ L.H.S = R.H.S

1194554_1390418_ans_20505b5b064c4b87b8e27641a07df253.jpg

Prove that: $\sec^{6}x-\tan^{6}x=1+3\sec^{2}x\times \tan^{2}x$

Now,

$\sec^{6}x-\tan^{6}x$

$=(\sec^{2}x)^3-(\tan^{2}x)^3$

$=\{\sec^2 x-\tan^2 x\}^3+3\sec^2 x.\tan^2 x(\sec^2 x-\tan^2 x)$

$=1+3\sec^2 x.\tan^2 x$ . [ Since

$\sec^2 x-\tan^2 x=1$ ]

$\dfrac{\cos A+\sin A}{\cos A-\sin A}=0$ find $A$

$\textbf{Simplifying the given expression according to given restraints}$

$\text{We have given,}$

$\Rightarrow \dfrac{\cos A+ \sin A}{\cos A -\sin A}=0$

$\Rightarrow \cos A+ \sin A=0$

$\Rightarrow \cos A=-\sin A$

$\Rightarrow\tan A=-1=\tan\dfrac{3\pi}{4}$

$\boldsymbol{\left[\because \tan A=\dfrac{\sin A}{\cos A}\right]}$

$\Rightarrow A=\dfrac{3\pi}{4}$

$\textbf{Hence value of A is}$

$\boldsymbol{\dfrac{3\pi}{4}}$

Solve the equation $(\cos x-\sin x)(2\tan x+2)=0$

$(\cos x -\sin x )(2\tan x+2)=0\\\cos x-\sin x=0\\\sin x=\cos x\\\tan x=1\\x=\dfrac{\pi}{4}\\2\tan x+2=0\\\tan x=-1\\x=\dfrac{3\pi}{4}$

Prove that:
$\dfrac{\sec A-\tan A}{\sec A+\tan A}=\left(\dfrac{\cos A}{1+\sin A}\right)^2$ .

$L.H.S=\frac{sec A- tan A}{sec A+tan A}$

Rationalizing the denominator by multiply and dividing with sec A+ tan A.

$\displaystyle \frac{sec A- tan A}{sec A+ tan A}\times \frac{sec A+tan A}{sec A+ tan A}=\frac{sec^{2}A-tan^{2}A}{(sec A+tan A)^{2}}$

$=\dfrac{1}{sec^{2}A+tan^{2}A+2.secA.tan A}$

$[\because sec^{2}A-tan^{2}A=1]$

$=\dfrac{1}{\dfrac{1}{cos^{2}A}+\dfrac{sin^{2}A}{cos^{2}A}+2.\dfrac{1}{cos A}.\dfrac{sin A}{cos A}}$

$\Rightarrow \dfrac{1}{\dfrac{1+sin^{2}A+2 sin A}{cos^{2}A}}$

$\Rightarrow \dfrac{cos^{2}A}{1+sin^{2}A+2.sin A}$

$=\dfrac{cos^{2}A}{(1+sin A)^{2}}$

$= R.H.S$

Find the principal solution: $2+\sqrt { 3 } sec{x}-4cosx=2\sqrt { 3 }$

$2+\sqrt{3}\sec{x}-4\cos{x}=2\sqrt{3}$

$\Rightarrow 2-4\cos{x}=2\sqrt{3}-\sqrt{3}\sec{x}$

$\Rightarrow 2\left(1-2\cos{x}\right)=\sqrt{3}\left(2-\sec{x}\right)$

$\Rightarrow \dfrac{1-2\cos{x}}{2-\dfrac{1}{\cos{x}}}=\dfrac{\sqrt{3}}{2}$

$\Rightarrow \dfrac{\cos{x}\left(1-2\cos{x}\right)}{2\cos{x}-1}=\dfrac{\sqrt{3}}{2}$

$\Rightarrow -cos{x}=\dfrac{\sqrt{3}}{2}$

$\Rightarrow cos{x}=\dfrac{-\sqrt{3}}{2}$ is in second and third quadrant.

$\therefore x=\pi-\dfrac{\pi}{3},\pi+\dfrac{\pi}{3}$

$\therefore x=\dfrac{3\pi-\pi}{3},\dfrac{3\pi+\pi}{3}$

$\therefore x=\dfrac{2\pi}{3},\dfrac{4\pi}{3}$

Prove the following:
$\dfrac{(\tan\theta+\sec\theta-1)}{\tan\theta-\sec\theta+1)}=\dfrac{1+\sin\theta}{\cos\theta}$ .

LHS

$\dfrac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +1}$

$\dfrac{\tan \theta +\sec \theta -(\sec^2 \theta -\tan^2 \theta) }{\tan \theta -\sec \theta +1}$

$\dfrac{\tan \theta +\sec \theta -(\sec \theta -\tan \theta)(\sec \theta +\tan \theta ) }{\tan \theta -\sec \theta +1}$

$\dfrac{\tan \theta +\sec \theta(1 -\sec \theta +\tan \theta) }{\tan \theta -\sec \theta +1}$

$\tan \theta +\sec \theta \\\dfrac{\sin \theta}{\cos \theta}+\dfrac 1{\cos \theta}\\\dfrac{1+\sin \theta}{\cos \theta}$

Prove that : $\left( {\sin \theta + \cos \theta } \right)\left( {\tan \theta + \cot \theta } \right) = \sec \theta + \text{cosec}\theta$

Given

$(sin \theta +cos \theta )(tan \theta +cot \theta )=sec \theta +cosec \theta$

$L.H.S= (sin \theta +cos \theta )(tan \theta +cot \theta )$

$=(sin \theta +cos \theta )(\dfrac{sin \theta }{cos \theta }+\dfrac{cos \theta }{sin \theta })$

$=(sin \theta +cos \theta )(\dfrac{sin^{2} \theta +cos^{2} \theta }{sin \theta .cos \theta })$

$=(sin \theta +cos \theta )\dfrac{1}{sin \theta .cos \theta }$

$=\dfrac{sin \theta +cos \theta }{sin \theta .cos \theta }$

$=\dfrac{sin \theta }{sin \theta cos \theta }+\dfrac{cos \theta }{sin \theta cos \theta }$

$=\dfrac{1}{cos \theta }+\dfrac{1}{sin \theta }$

$=sec \theta +cosec \theta$

$=R.H.S$

$\therefore (sin \theta +cos \theta )(tan \theta +cot \theta )=sec \theta +cosec \theta$

If $(a\sin A)=b\cos A$ and $a\sin^{3}A+b\cos^{3}A=\sin A\cos A$ , then prove $a^{2}+b^{2}=1$

$a\sin A(\sin^2A)+b\cos A(\cos^2A)\sin A\cos A$

$a\sin A(1)=\sin A\cos A$

$\Rightarrow a=\cos A$

$\Rightarrow b=\sin A$

$\Rightarrow a^2+b^2=\sin^2A+\cos^2A=1$ .

1216086_1396869_ans_e6c6624fe9894593bd68d107eab9b4cd.jpg

If $\sec \theta + \tan \theta = p$ , then find the value $\text{cosec} \theta$ .

$sec\theta +tan \theta =P$ _______ (1)

$sec^{2}\theta -tan^{2}\theta =1$

$(sec \theta +tan \theta )(sec \theta - tan \theta )=1$

$sec\theta -tan \theta = 1/P$ _______ (2)

Subtracting (2) from (1), we get

$2 tan \theta = P-\dfrac{1}{P}$

$\therefore cot \theta =\dfrac{2P}{P^{2}-1}$

Now,

$cosec^{2}\theta -cot^{2}\theta =1$

$cosec^{2}\theta =1+cot^{2}\theta$

$\Rightarrow cosec^{2}\theta =1+(\dfrac{2P}{P^{2}-1})^{2}$

$\Rightarrow cosec^{2}\theta =1+\dfrac{4P^{2}}{(P^{2}-1)^{2}}$

$\Rightarrow cosec^{2}\theta =(P^{4}-2P^{2}+1+4P^{2})/(P^{2}-1)^{2}$

$\Rightarrow cosec^{2}\theta =\dfrac{P^{2}+1}{P^{2}-1}$

1201488_1413678_ans_cf88ebe7c96a492d99942670a1fd3b66.jpg

If a line makes angles $\alpha, \beta, \gamma$ with positive direction co-ordinate axes then prove that $\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=2$

The angles made by line with axis is

$\alpha ,\beta,\gamma$

For a line

$\cos ^2\alpha +\cos ^2\beta +\cos ^2\gamma =1\\1-\sin ^2 \alpha +1-\sin ^2\beta+1-\sin ^2\gamma=1\\3-(\sin ^2\alpha +\sin ^2\beta+\sin ^2\gamma)=1\\\sin ^2\alpha +\sin ^2\beta+\sin ^2\gamma=2$

Prove that $2{\sec ^2}\theta - {\sec ^4}\theta - 2\cos e{c^2}\theta + \cos e{c^4}\theta = {\cot ^4}\theta - {\tan ^4}\theta$

Given

$2sec^{2}\theta -sec^{4}\theta -2cosec^{2}\theta +cosec^{4}\theta$

$=2(1+tan^{2}\theta )-(sec^{4}\theta )-2(1+cot^{2}\theta )+cosec^{4}\theta$

$=2+2tan^{2}\theta -(sec^{2}\theta )^{2}-2-2cot^{2}\theta +(1+cot^{2}\theta )^{2}$

$=2tan^{2}\theta -(1+tan^{2}\theta )^{2}-2.cot^{2}\theta +(1+cot^{2}\theta )^{2}$

$=2tan^{2}\theta -2cot^{2}\theta =(1+2tan^{2}\theta +tan^{4}\theta )+(1+2cot^{2}\theta +cot^{4}\theta )$

$=2.tan^{2}\theta -2cot^{2}\theta -1-2tan^{2}\theta -tan^{4}\theta +1+2.cot^{2}\theta +cot^{4}\theta$

$=cot^{4}\theta -tan^{4}\theta$

$\therefore L.H.S = R.H.S$

Prove that $\text{cosec}^6\theta = {\cot ^6}\theta + 3{\cot ^2}\theta .\text{cosec}^2\theta + 1$

$cosec^{6}\theta =(cosec^{2}\theta )^{3}.$

$cosec^{2}\theta =cot^{2}\theta+1 ----(1)$

$(cosec^{2}\theta)^{3}=(cot^{2}\theta +1)^{3}$

Since

$(x+y)^{3}=x^{3}+y^{3}+3xy(x+y),$ we have

$cosec^{6}\theta =(cot^{2}\theta )^{3}+(1) ^{3}(cot^{2}\theta )(1)(cot^{2}\theta +1)$

$\Rightarrow cosec^{6}\theta = cot^{6}\theta +1+3.cot^{2}.cosec^{2}\theta$

$\Rightarrow cosec^{6}\theta = cot^{6}\theta +3.cot^{2} cosec^{2}\theta+1$

$\therefore$ L.H.S=R.H.S

Prove that $\displaystyle \frac{{\sin \theta }}{{1 - \cos \theta }} + \frac{{\tan \theta }}{{1 + \cos \theta }} = \cot \theta + \sec \theta .\text{cosec}\theta$

1212905_1435665_ans_ee75d964966447418e78fd5cd6aee4aa.jpg

Evaluate :
$\dfrac { cosA-sinA+1 }{ cosA+sinA-1 } =cosecA+cotA$ .

1308730_1435818_ans_83ced7e0185a498588ab2434800b29f1.JPG

$\sin (A+B)=\dfrac{\surd{3}}{2}$ and $\cos (A-B)=\dfrac{\surd{3}}{2}$ , where $(A+B)$ and $(A-B)$ are acute angles. Find the values of $A$ and $B$ $A,B <90^{0}$

$\sin{\left(A+B\right)}=\dfrac{\sqrt{3}}{2}$

$\Rightarrow A+B=\dfrac{\pi}{3}$ .......

$(1)$

$\cos{\left(A-B\right)}=\dfrac{\sqrt{3}}{2}$

$\Rightarrow A-B=\dfrac{\pi}{6}$ ......

$(2)$

Adding eqns

$(1)$ and

$(2)$ we get

$A+B+A-B=\dfrac{\pi}{3}+\dfrac{\pi}{6}$

$\Rightarrow 2A=\dfrac{2\pi+\pi}{6}=\dfrac{3\pi}{6}=\dfrac{\pi}{2}$

$\therefore A= \dfrac{\pi}{4}$

Substitute

$A= \dfrac{\pi}{4}$ in

$(1)$ we get

$A+B=\dfrac{\pi}{3}$

$\Rightarrow B=\dfrac{\pi}{3}-A$

$\Rightarrow B=\dfrac{\pi}{3}-\dfrac{\pi}{4}=\dfrac{\pi}{12}$

$\therefore A=\dfrac{\pi}{4}$ and

$B=\dfrac{\pi}{12}$

Prove that $\dfrac{{\sin \theta - 2\,{{\sin }^3}\theta }}{{2\,{{\cos }^3}\theta - \cos \,\theta }} = \tan \theta$

$\dfrac{sin\theta -2sin^{3}\theta }{2cos^{3}\theta -cos\theta }=tan\theta$

$L.H.S=\dfrac{sin\theta -2sin^{3}\theta }{2.cos^{3}\theta -cos\theta }$

$=\dfrac{sin\theta(1-2sin^{2}\theta )}{cos\theta (2cos^{2}\theta -1)}$

$=\dfrac{sin\theta [1-2(1-cos^{2}\theta )]}{cos\theta (2.cos^{2}\theta -1)}$

$=\dfrac{sin\theta (1-2+2cos^{2}\theta ]}{cos\theta (2cos^{2}\theta -1)}$

$=\dfrac{sin\theta (2cos^{2}\theta -1)}{cos\theta (2cos^{2}\theta -1)}$

$=\dfrac{sin\theta }{cos\theta }$

$=tan\theta$

$=R.H.S$

$\therefore L.H.S=R.H.S$

Hence proved.

Find the values of $\theta$ ,( in degrees) if $\sin (\theta +24)=\cos \theta$ , where $(\theta+24)^{0}$ is an acute angle .

$\sin{\left(\theta+{24}^{\circ}\right)}=\cos{\theta}$

$\Rightarrow \sin{\left(\theta+{24}^{\circ}\right)}=\sin{\left({90}^{\circ}-\theta\right)}$ since

$\sin{\left({90}^{\circ}-\theta\right)}=\cos{\theta}$

$\Rightarrow \theta+{24}^{\circ}={90}^{\circ}-\theta$

$\Rightarrow 2\theta={90}^{\circ}-{24}^{\circ}={66}^{\circ}$

$\therefore \theta=\dfrac{{66}^{\circ}}{2}={33}^{\circ}$

Prove $\displaystyle \frac{{\tan \theta }}{{1 - \cot \theta }} + \frac{{\cot \theta }}{{1 - \tan \theta }} = 1 + cosec \theta. \sec \theta$

1212863_1436309_ans_10b21e496edf4f41a02e29dd303daef3.jpg

If $\sec \theta - \tan \theta = x$ , then show that $\sec \theta + \tan \theta = \dfrac{1}{x}$ and hence find the values of $\cos \theta$ and $\sin \theta$ .

Given:

$\sec\theta -\tan\theta =x$ …………….

$(1)$

To prove:

$\sec\theta +\tan\theta =1/x$

By trigonometric identity

$\Rightarrow \sec^2\theta -\tan^2\theta =1$

$\Rightarrow (\sec\theta +\tan\theta)(\sec\theta -\tan\theta)=1$

Put value of

$(1)$

$\Rightarrow x.(\sec\theta +\tan\theta)=1$

$\therefore \sec \theta +\tan\theta =1/x$

Now,

$\sec\theta -\tan\theta =x$ …………….

$(1)$

$\sec\theta +\tan\theta =1/x$ ………….

$(2)$

Add both equations

$\sec\theta -\tan\theta +\sec\theta +\tan\theta =x+\dfrac{1}{x}$

$\Rightarrow 2\cdot \sec\theta =\dfrac{x^2+1}{x}$

$\Rightarrow \sec\theta =\dfrac{x^2+1}{2x}$

$\Rightarrow \dfrac{1}{\cos\theta}=\dfrac{x^2+1}{2x} \left[\cos\theta =\dfrac{1}{\sec\theta}\right]$

$\Rightarrow \cos\theta =\dfrac{2x}{x^2+1}$

using trigonometric identify

$\Rightarrow \sin\theta =\sqrt{1-\cos^2\theta}$

$\Rightarrow \sin\theta =\sqrt{1-(2x/x^2+1)^2}$

$\Rightarrow \sin\theta =\sqrt{1-\dfrac{4x^2}{(x^2+1)^2}}$

$\Rightarrow \sin\theta =\sqrt{\dfrac{(x^2+1)^2-4x^2}{(x^2+1)^2}}$

$\Rightarrow \sin\theta =\sqrt{\dfrac{x^4+1-2x^2}{(x^2+1)^2}}$

$\Rightarrow \sin\theta =\sqrt{\dfrac{(x^2-1)^2}{(x^2+1)^2}}$

$\Rightarrow \sin\theta =\sqrt{\left(\dfrac{x^2-1}{x^2+1}\right)^2}$

$\therefore \sin\theta =\dfrac{x^2-1}{x^2+1}$

$\therefore \cos\theta =\dfrac{2x}{x^2+1} : \sin\theta =\dfrac{x^2-1}{x^2+1}$ .

Prove $\displaystyle \frac{{1 + \cos A}}{{\sin A}} + \frac{{\sin A}}{{1 + \cos A}} = 2\text{cosec} \ A$

L.H.S.

$=\dfrac{1+\cos A}{\sin A}+\dfrac{\sin A}{1+\cos A}$

$=\dfrac{(1+\cos A)(1+\cos A)+\sin A\sin A}{\sin A(1+\cos A)}$

$=\dfrac{(1+\cos A)^2+\sin^2A}{(1+\cos A)\sin A}$

$=\dfrac{1+\cos^2A+2\cos A+\sin^2A}{(1+\cos A)\sin A}$

$=\dfrac{1+2\cos A+1}{(1+\cos A)\sin A}$

$=\dfrac{2+2\cos A}{(1+\cos A)\sin A}$

$=\dfrac{2(1+\cos A)}{(1+\cos A)\sin A}$

$=2\dfrac{1}{\sin A}$

$=2 cosec A$

$=$ R.H.S

Hence proved.

Evaluate : $\dfrac { \cos x } { ( 1 + \sin x ) ^ { 2 } ( 2 + \sin x ) }$

1231636_1503419_ans_3712bb87aff14cbeb4d0d9786ae11dc4.JPG

Prove that $\displaystyle \frac{{{\mathop{\rm tanA}\nolimits} }}{{1 + \sec A}} - \frac{{\tan A}}{{1 - \sec A}} = 2\text{cosec}A$

$\dfrac{\tan A}{1+\sec A}-\dfrac{\tan A}{1-\sec A}=2\cdot cosec A$

L.H.S

$\dfrac{\tan A}{1+\sec A}-\dfrac{\tan A}{1-\sec A}$

$=\dfrac{\tan A(1-\sec A)-\tan A(1+\sec A)}{1-\sec^2A}$

$=\dfrac{\tan A(1-\sec A-1-\sec A)}{-\tan^2A}$

$=\dfrac{-2\sec A}{-\tan A}$

$=\dfrac{2\cdot 1/cos A}{\sin A/cos A}$

$=\dfrac{2}{\sin A}$

$=2\cdot cosec A$

$=$ R.H.S

$\therefore$ L.H.S

$=$ R.H.S.

Prove that $\displaystyle \frac{{\tan \theta - \cot \theta }}{{\sin \theta \cos \theta }} = {\tan ^2}\theta - {\cot ^2}\theta$

We have,

$\dfrac{\tan \theta-\cot \theta}{\sin \theta.\cos \theta}$

$=\sec^2\theta-cosec^2 \theta$

$=1+\tan^2\theta-1-\cot^2 \theta$

$=\tan^2 \theta-\cot^2 \theta$ .

Solve: $\left(\cos{x}-\sin{x}\right)\left(2\tan{x}+\dfrac{1}{\cos{x}}\right)+2=0$

$\left(\cos{x}-\sin{x}\right)\left(2\tan{x}+\dfrac{1}{\cos{x}}\right)+2=0$

Let

$t=\tan{\dfrac{x}{2}}$

$=\left\{\dfrac{1-{\tan}^{2}{\frac{x}{2}}}{1+{\tan}^{2}{\frac{x}{2}}}-\dfrac{2\tan{\dfrac{x}{2}}}{1+{\tan}^{2}{\frac{x}{2}}}\right\}\left\{\dfrac{4\tan{\frac{x}{2}}}{1-{\tan}^{2}{\frac{x}{2}}}+\dfrac{1+{\tan}^{2}{\frac{x}{2}}}{1-{\tan}^{2}{\frac{x}{2}}}\right\}+2=0$

$\left(\dfrac{1-{t}^{2}}{1+{t}^{2}}-\dfrac{2t}{1-{t}^{2}}\right)\left(\dfrac{4t}{1-{t}^{2}}+\dfrac{1+{t}^{2}}{1-{t}^{2}}\right)+2=0$

$\Rightarrow \dfrac{1-{t}^{2}-2t}{1+{t}^{2}}\times\dfrac{4t+1+{t}^{2}}{1-{t}^{2}}=0$

$\Rightarrow \dfrac{-\left({t}^{2}+2t-1\right)\left({t}^{2}+4t+1\right)}{1-{t}^{4}}=0$

$\Rightarrow -\dfrac{{t}^{4}+4{t}^{3}+{t}^{2}+2{t}^{3}+8{t}^{2}+2t-{t}^{2}-4t-1}{1-{t}^{4}}=0$

$\Rightarrow -\dfrac{{t}^{4}+6{t}^{3}+8{t}^{2}-2t-1}{1-{t}^{4}}=0$

$\Rightarrow \dfrac{{t}^{4}+6{t}^{3}+8{t}^{2}-2t-1}{1-{t}^{4}}=0$

$\Rightarrow \dfrac{{t}^{4}-1}{1-{t}^{4}}+\dfrac{2t\left(3{t}^{2}+4t-1\right)}{1-{t}^{4}}=0$

$\Rightarrow \,-1+\dfrac{2t\left(3{t}^{2}+4t-1\right)}{1-{t}^{4}}=0$

$\Rightarrow \,\dfrac{2t\left(3{t}^{2}+4t-1\right)}{1-{t}^{4}}=1$

$\Rightarrow \,2t\left(3{t}^{2}+4t-1\right)=1-{t}^{4}$

$\Rightarrow 6{t}^{3}+8{t}^{2}-2t-1+{t}^{4}=0$

$\Rightarrow {t}^{4}+6{t}^{3}+8{t}^{2}-2t-1=0$

Let

$t=\dfrac{\pm 1}{\sqrt{3}}\Rightarrow {\left(\dfrac{\pm 1}{\sqrt{3}}\right)}^{4}+6{\left(\dfrac{\pm 1}{\sqrt{3}}\right)}^{3}+8{\left(\dfrac{\pm 1}{\sqrt{3}}\right)}^{2}-2\left(\dfrac{\pm 1}{\sqrt{3}}\right)-1=0$

Hence its roots are

${t}_{1}=\dfrac{1}{\sqrt{3}}$ and

${t}_{2}=\dfrac{-1}{\sqrt{3}}$

$\Rightarrow \dfrac{x}{2}=n\pi\pm\dfrac{\pi}{6}$

$\Rightarrow \,x=2n\pi\pm\dfrac{\pi}{3}$ is required solution.

Prove that $\dfrac { \cos A+ \sin A-1 }{ \cos A-\sin A+1 } =\dfrac { 1 }{ \text{cosec}A+\cot A }$ , using the identity ${ \text{cosec }}^{ 2 }A-{ \cot }^{ 2 }A=1.$

We have,

$\dfrac{cosA + sinA - 1}{cosA - (sinA - 1)} \times \dfrac{cosA + (sinA - 1)}{cosA + (sinA + 1)}$

$\Rightarrow \dfrac{cos^{2}A + (sinA - 1)^{2} + 2cosA(sin-1)}{cos^{2}A - (sinA - 1)^{2}}$

$\Rightarrow \dfrac{cos^{2}A + sin^{2} A + 1 2 sinA + 2cosA sinA - 2cosA}{cos^{2}A - sin^{2}A - 1 + 2sinA}$

$= \dfrac{2(1-sinA) + 2cosA(sinA-1)}{-2sin^{2}A+2sinA}$

$= \dfrac{2(1-cosA)(1-sinA)}{-2sinA (sinA-1)} = \dfrac{(1-cosA) \times (1+cosA)}{sinA (1+cosA)}$

$\Rightarrow \dfrac{sin^{2}A}{sinA(1+cosA)} = \dfrac{1}{\dfrac{1}{sinA} + \dfrac{cosA}{sinA}}$

$\Rightarrow \dfrac{1}{cosecA + cotA}$

Prove $\left (\dfrac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left (\dfrac{1 - \tan A}{1 - \cot A}\right)^2$

$LHS =$

$- \dfrac{sec^{2}A}{cosec^{2}A}=\dfrac{+sin^{2}A}{cos^{2}A}=tan^{2}A$

$RHS= \left ( \dfrac{1-tanA}{1-cotA} \right )^{2}$

$=\left [ \left ( \dfrac{1-tanA}{tanA-1} \right ) tanA\right ]^{2}$

$= tan^{2}A$

$\therefore LHS= RHS$

If $\cos \left( {2{{\tan }^{ - 1}}x} \right) = \dfrac{1}{2},\;then\;the\;value\;of\;x\;is\;........$

Prove that : $\dfrac{{1 + \cos A + \sin A}}{{1 + \cos A - \sin A}} = \dfrac{{1 + {\mathop{\rm sinA}\nolimits} }}{{\cos A}}$

We have,

$\displaystyle \frac{1+cos A+ sin A}{1+ csos A - sin A}$

$=\dfrac{1+ cos A+ sin A}{1+ cos A - sin A}\times \dfrac{(1+ cos A)+ sin A}{(1+ cos A)+ sin A}$

$=\dfrac{((1+cos A)+sin A)^{2}}{(1+ cos A)^{2}- sin^{2}A}$

$=\dfrac{(1+ cos A)^{2}+sin^{2}A+2(1+cos A)sin A}{1+ cos^{2}A+2 cos A-1+ cos^{2}A}$

$=\dfrac{1+cos^{2}A+sin^{2}A+2 cos A+2 sin A+2 sin A cos A}{2 cos^{2}A+2 cos A}$

$=\dfrac{1+ cos^{2}A+sin^{2}A+2 cos A+2 sin A+2 sin A cos A}{2 cos^{2}A+2 cos A}$

$\dfrac{1+1+2 cos A+2 sin A+2 sin A cos A}{2 cos A (1+ cos A)}$

$\dfrac{2+2 cos A+2 sin A+2 sin A cos A}{2 cos A (1+ cos A)}$

$\dfrac{1+ cos A+sin A+sin A cos A}{cos A (1+ cos A)}$

$=\dfrac{1+ sin A + cos A (1+ sin A)}{cos A (1+cos A)}$

$=\dfrac{(1+ sin A)(1+ cos A)}{cos A (1+cos A)}$

$=\dfrac{1+ sin A}{cos A}$

Hence proved.

Find the value of $\theta$ ,( in degrees ) if, $\dfrac{{\cos \theta }}{{1 - \sin \theta }} + \dfrac{{\cos \theta }}{{1 + \sin \theta }} = 4;\,\,\theta \le 90^\circ$

1211953_1506033_ans_81a14a1f9f6b49ca9dd4c5e8957db901.jpg

Prove that : $\dfrac{{\text{cosec}\theta - 1}}{{\text{cosec}\theta + 1}} = \dfrac{{{{\cos }^2}\theta }}{{{{\left( {1 + \sin \theta } \right)}^2}}}$

Given,

$\dfrac{cosec \theta -1}{cosec \theta +1}$

$=\dfrac{\dfrac{1}{sin \theta }-1}{\dfrac{1}{sin \theta }+1}$

$=\dfrac{\dfrac{1-sin\theta }{sin \theta }}{\dfrac{1+ sin \theta }{sin \theta }}$

$=\dfrac{1-sin \theta }{1+ sin \theta }\times \dfrac{1+sin\theta }{1+ sin \theta }$

$=\dfrac{1- sin^{2}\theta }{(1+sin \theta )}$

$=\dfrac{cos^{2}\theta }{(1+ sin \theta )^{2}}$

If ${ sin }^{ 2 }\theta +5{ cos }^{ 2 }\theta =4,$ then find $\theta$ and hence prove that $sec\theta +cosec\theta =2+\dfrac { 2 }{ \sqrt { 3 } }$

Given,

${ sin }^{ 2 }\theta +5{ cos }^{ 2 }\theta =4,$

$\Rightarrow 1-{\cos}^{2}{\theta}+5{\cos}^{2}{\theta}=4$

$\Rightarrow 4\cos^{2}\theta$ =

$3$

Solving we get

$\cos\theta= \dfrac{\sqrt{3}}{2}$

$\theta=30^{\circ}$

Now

$\sec\theta + cosec\theta$ =

$\sec(30)^{\circ}+cosec(30)^{\circ}=$

$\dfrac{2}{\sqrt{3}}+2$

Prove the following identities :
(i) $( \sin \theta + \csc \theta ) ^ { 2 } = \sin ^ { 2 } \theta + \csc ^ { 2 } \theta + 2$
(ii) $\cos ^ { 4 } \theta - \sin ^ { 4 } \theta = 1 - 2 \sin ^ { 2 } \theta$

$( \sin \theta + \csc \theta ) ^ { 2 } = \sin ^ { 2 } \theta + \csc ^ { 2 } \theta + 2$ (ii)

$\cos ^ { 4 } \theta - \sin ^ { 4 } \theta = 1 - 2 \sin ^ { 2 } \theta.$

${\left(sin\theta+csc\theta\right)}^{2}={sin}^{2}\theta+{csc}^{2}\theta+2sin\theta csc\theta$

${cos}^{4}\theta-{sin}^{4}\theta=\left({cos}^{2}\theta-{sin}^{2}\theta\right)\left({cos}^{2}\theta+{sin}^{2}\theta\right)$

$={sin}^{2}\theta+{csc}^{2}\theta+2sin\theta\dfrac{1}{sin\theta}$

$={cos}^{2}\theta(1)$

$={sin}^{2}\theta+{csc}^{2}\theta+2$

$=1-2{sin}^{2}\theta$

Find the principal solution of $\sqrt 3 \sec x + 2 = 0$

1972423_1505428_ans_ec38f2dfd0b143f1aa89b8427172a66e.jpeg

Find the magnitude, in radians and degrees, of the interior angle of
(1) a regular pentagon,
(2) a regular heptagon,
(3) a regular octagon,
(4) a regular duodecagon,
and (5) a regular polygon of $17$ sides.

1643557_1508752_ans_71da90c02f044efea0a01240e216c87d.PNG

If $\tan \theta + \sin \theta = m$ and $\tan \theta - \sin \theta = n ,$ then
(i) $m ^ { 2 } - n ^ { 2 } = 4 m n$
(ii) $m ^ { 2 } + n ^ { 2 } = 4 m n$
(iii) $m ^ { 2 } - n ^ { 2 } = m ^ { 2 } + n ^ { 2 }$
(iv) $m ^ { 2 } - n ^ { 2 } = 4 \sqrt { m n }$

1870853_1561943_ans_d672948ed1bc4edebe6a467c7b0defb7.jpg

Prove that $\dfrac{\tan A}{1+ \sec A}-\dfrac{\tan A}{1 -\sec A}=2\text{cosec}A$

$LHS=\dfrac{tan A}{1+ secA}-\dfrac{tan A}{1- sec A}$

$= \dfrac{tan A(1- sec A)- tan A (1+ sec A)}{1- sec^{2}A}$

$=\dfrac{tan A(1-sec A - 1 - sec A)}{-(sec^{2}A-1)}$

$=\dfrac{tan A (-2 sec A)}{- tan^{2}A}$

$=\dfrac{+2/ cos A}{+ sin A/ cos A}$

$=\dfrac{2}{sin A}= 3 cosec A$

Prove that : $sec^{4}\Theta (1-sin^{4}\Theta )-2tan^{2}\Theta =1$

As we know that

$a^2-b^2=(a-b)(a+b)$

$LHS=\text{sec}^4 \theta(1-\sin^4 \theta)-2\tan^2\theta$

$=\text{sec}^4 \theta(1-\sin^2 \theta)(1+\sin^2 \theta)-2\tan^2 \theta$

$=\text{sec}^4 \theta\cos^2 \theta(1+\sin^2 \theta)-2\tan^2 \theta$

$=\text{sec}^2 \theta(1+\sin^2 \theta)-2\tan^2 \theta$

$=\text{sec}^2 \theta+\text{sec}^2 \theta\sin^2 \theta-2\tan^2 \theta$

$=\text{sec}^2 \theta+\tan^2 \theta-2\tan^2 \theta$

$=\text{sec}^2 \theta-\tan^2 \theta$

$=1$

$=RHS$

Solve the equation $sinx+cosx=1$

$\sin{x}+\cos{x}=1$

$\Rightarrow {\left(\sin{x}+\cos{x}\right)}^{2}=1$ by squaring both sides

$\Rightarrow {\sin}^{2}{x}+{\cos}^{2}{x}+2\sin{x}\cos{x}=1$

$\Rightarrow 1+2\sin{x}\cos{x}=1$ since

${\sin}^{2}{x}+{\cos}^{2}{x}=1$

$\Rightarrow 2\sin{x}\cos{x}=0$

$\Rightarrow \sin{2x}=0$

$\Rightarrow 2x=n\pi$

$\therefore x=2n\pi+\dfrac{\pi}{2},2n\pi$ for

$n\in Z$

$\frac{sinA-cosA+1}{sinA+cosA-1}=\frac{1}{secA-tanA}$

LHS

$= \dfrac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1}=\dfrac{\tan \theta - 1 + \sec \theta}{\tan \theta + 1 - \sec \theta}$

$= \dfrac{(\tan \theta + \sec \theta)-1}{(\tan \theta - \sec \theta)+1} = \dfrac{\{(\tan \theta + \sec \theta)-1\} (\tan \theta - \sec \theta)}{\{(\tan \theta - \sec \theta)+1\} (\tan \theta - \sec \theta)}$

$= \dfrac{(\tan^2 \theta - \sec^2 \theta)- (\tan \theta - \sec \theta)}{\{\tan \theta - \sec \theta + 1\}(\tan \theta -\sec \theta)}$

$= \dfrac{-1}{\tan \theta - \sec \theta}$

$[\because \tan^2 \theta - \sec^2 \theta = -1]$

$= \dfrac{1}{\sec \theta - \tan \theta}$

Fin an acute angle $\Theta$ , when
$\frac{cos\Theta -sin\Theta }{cos\Theta +sin\Theta }=\frac{1-{\sqrt{3}}}{1+\sqrt{3}}$ .

Given

$\dfrac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}=\dfrac{1-\sqrt{3}}{1+\sqrt{3}}$

$\implies \dfrac{1-\frac{\sin \theta}{\cos \theta}}{1+\frac{\sin \theta}{\cos \theta}}=\dfrac{(1-\sqrt{3})^2}{(1+\sqrt{3})(1-\sqrt{3})}$

$\implies \dfrac{1-\tan \theta}{1+\tan \theta}=\dfrac{1+3-2\sqrt{3}}{1-3}$

$\implies \dfrac{\tan \frac{\pi}{4}-\tan \theta}{1+\tan \frac{\pi}{4}\tan \theta}=\dfrac{4-2\sqrt{3}}{-2}$

$\implies \tan \bigg(\frac{\pi}{4}-\theta\bigg)=\sqrt{3}-2$

$\implies \dfrac{\pi}{4}-\theta=-\dfrac{\pi}{12}$

$\implies \theta=\dfrac{\pi}{4}+\dfrac{\pi}{12}$

$\implies \theta=\dfrac{\pi }{3}$

Find the degree measure of ${\dfrac{2\pi}{15}}^{c}$

We have

${\pi}^{c}={180}^{\circ}$

$\therefore\,{1}^{c}={\dfrac{180}{\pi}}^{\circ}$

${\dfrac{2\pi}{15}}^{c}={\left(\dfrac{2\pi}{15}\times \dfrac{180}{\pi}\right)}^{\circ}={24}^{\circ}$

Solve, by factorising, the equation $2\cos \theta\sin\theta-2\cos\theta-\sin\theta+1=0$ for $0\le\theta\le \pi$

$2\cos \theta \sin \theta -2\cos \theta -\sin \theta +1=0$

$2\cos \theta (\sin \theta -1)-1(\sin \theta -1)=0$

$(2\cos\theta-1)(\sin\theta-1)=0$

$\cos\theta=\dfrac 12$ or

$\sin\theta=1$

$\theta=\dfrac \pi 3$ or

$\dfrac \pi 2$

Find the principal solution of the following equation:
$co\sec { x } =2$

1981694_1710015_ans_be3c63f053f84b0497bb7d2d0de4685f.jpg

Solve the following equation:
$2^{\sin^{2}x}+2^{\cos ^{2}x}=2\sqrt 2$

$2^{\sin^{2}x}+2^{\cos ^{2}x}=2\sqrt 2$

$\Rightarrow 2^{\sin^{2}x}+2^{1-\sin ^{2}x}=2\sqrt 2$

$\Rightarrow \ y+\dfrac {2}{y}=2\sqrt 2$ , where

$y=2^{\sin ^2 x}$

$\Rightarrow \ y^2-2\sqrt 2 y+2=0$

$\Rightarrow (y-\sqrt 2)^2 =0$

$\Rightarrow y=\sqrt 2$

$\Rightarrow 2^{\sin^2 x}=2^{1/2}$

$\Rightarrow \sin^2 x=\dfrac {1}{2}=\sin^2\dfrac {\pi}{4}$

Find the principal solution of the following equation:
$\sqrt { 3 } co\sec { x } +2=0$

$\sqrt { 3 } co\sec { x } +2=0$

$co\sec { x } =\dfrac{-2}{\sqrt3}$

$\cfrac { 4\pi }{ 3 } ,\cfrac { 5\pi }{ 3 }$

Find the principal solution of the following equation:
$\sqrt { 2 } \cos { x } +1=0$

1981697_1710022_ans_cfbe7f8b502246fc8b2a53deddcdbcdb.jpg

Find the principal solution of the following equation:
$\sin { x } =\cfrac { -1 }{ 2 }$

1981696_1710021_ans_13a605e1bc064ec5bf07e30f6f2cd716.jpg

If $\tan x=\dfrac {b}{a}$ , then find the value of $\sqrt {\dfrac {a+b}{a-b}}+ \sqrt {\dfrac {a-b}{a+b}}$

We have,

$\tan x=\dfrac {b}{a}$

$\therefore \ \sqrt {\dfrac {a+b}{a-b}}+\sqrt {\dfrac {a-b}{a+b}}$

$=\sqrt {\dfrac {1+\dfrac {b}{a}}{1-\dfrac {b}{a}}}+\sqrt {\dfrac {1-\dfrac {b}{a}}{1+\dfrac {b}{a}}}$

$=\sqrt {\dfrac {1+\tan x}{1-\tan x}}+\sqrt {\dfrac {1-\tan x}{1+\tan x}}$

$=\sqrt {\dfrac {\cos x+\sin x}{\cos x-\sin x}}+\sqrt {\dfrac {\cos x-\sin x}{\cos x+\sin x}}$

$=\dfrac {\cos x+\sin x +\cos x-\sin x}{\sqrt {\cos^2x-\sin^2 x}}=\dfrac {2\cos x}{\cos^2x-\sin^2x}$

Prove that :
$\cos 2\alpha \cos 2\beta +\sin^2 (\alpha -\beta)-\sin^2 (\alpha +\beta)=\cos 2(\alpha +\beta)$

LHS :

$\cos 2\alpha \cos 2 \beta +\sin^2 (\alpha -\beta)-\sin^2 (\alpha +\beta)$

$=\cos 2\alpha \cos 2\beta +\sin (\alpha -\beta +\alpha +\beta)\sin (\alpha -\beta -\alpha -\beta)$

$=\cos 2\alpha \cos 2\beta -\sin 2\alpha \sin 2\beta$

$=\cos (2\alpha +2\beta)$ = RHS

Express in terms of a right angle the angles
$60^{\circ}$ .
$75^{\circ} 15'$ .
$63^{\circ} 17'25''$ .
$130^{\circ} 30'$ .
$210^{\circ} 30'30''$ .
$370^{\circ} 20'48''$ .

1554436_1728848_ans_50fac23b4570498fa2fb33a6df0fc3fd.png

Mark the position of the revolving line when it has traced out the following angles:
$\dfrac {4}{3}$ right angle.
$3\dfrac {1}{2}$ right angles.
$13\dfrac {1}{3}$ right angles.
$120^{\circ}$
$315^{\circ}$
$745^{\circ}$
$1185^{\circ}$
$150^{g}$
$420^{g}$
$875^{g}$ .

$1.$

$\dfrac{4}{3}$ right angle

$=$

$1\dfrac{1}{3}$ rt. angle

$=(90^o+30^o)$

Here, since the revolving line makes an angle more than one and less than two right angles with the original line, hence it will be in the second quadrant.

$2.$

$3\dfrac{1}{2}$ right angle

$=3$ right angle

$+45^o$

Here, since revolving line makes an angle greater than three and less than four right angles with the given line, hence it will lie in the fourth quadrant.

$3.$

$13\dfrac{1}{3}$ right angle

$=12$ right angle

$+1\dfrac{1}{3}$ right angle.

Here, since the revolving line makes three complete revolutions and

$1\dfrac{1}{3}$ right angle with the base, hence revolving line is in second quadrant.

$4.$

$120^o=90^o+30^o$

Here, since revolving line makes an angle greater than one right angle and less than

$2$ right angles, hence it will lie in the fourth quadrant.

$5.$

$315^o=270^o+45^o=3$ right angle

$+45^o$

Here, since the revolving line makes an angle greater than three and less than four right angles with the given straight line, hence it will lie in the fourth quadrant.

$6.$

$745^o=2\times 360^o+25^o$

Here, since revolving makes two complete revolutions and an angle of

$25^o$ from its original position, hence in the final position revolving line will lie in the first quadrant.

$7.$

$1185^o=3\times 360^o+(90^o+15^o)$

Here, since revolving line makes two complete revolutions and an angle of

$15^o$ more than a right angle with its original position, hence it will lie in the second quadrant.

$8.$

$150^g=\dfrac{150}{100}$ right angles

$=1\dfrac{1}{2}$ right angles

Here, since revolving line here makes an angle of

$\dfrac{1}{2}$ right angle more than a right angle, hence revolving line will lie in the second quadrant.

$9.$

$420^g=\dfrac{420}{100}$ right angle

$4\dfrac{1}{5}$ right angles.

Here, since revolving line makes one complete revolution and an angle of

$18^o,$ hence in the final position the revolution line will be found in the first quadrant.

$10.$

$875^g=\dfrac{875}{100}$ right angles

$=8.75$ right angles.

Here, since revolving line makes two complete revolutions and an angle

$67.5^o$ , hence the revolving line will lie in the first quadrant.

Express in grades, minutes, and seconds the angles
$30^{\circ}$
$81^{\circ}$
$138^{\circ} 30'$
$35^{\circ} 47'15''$
$235^{\circ} 12'36''$
$475^{\circ} 13'48''$ .

$1.$

$30^o=30\times\dfrac{10^g}{9}=33^g33'33.3"$

$2.$

$81^o=81\times\dfrac{10^g}{9}=90^g$

$3.$

$138^o30'=138\dfrac{1}{2}\times\dfrac{10^g}{9}=\dfrac{277}{2}\times\dfrac{10^g}{9}$

$=\dfrac{2770}{18}=153\dfrac{16"}{18}=153^g88'88.8"$

$4.$

$35^o47'15"=35^o\,47\dfrac{1}{4}'=35^o+\dfrac{189\times 1^o}{4\times 60}=35\dfrac{63^o}{80}$

$=\dfrac{2863}{80}\times\dfrac{10^g}{9}$

$=\dfrac{2863}{72}=39^g76'38.8"$

$5.$

$235^o12'36"=235^o\,12\dfrac{3}{5}=235^o+\dfrac{63^o}{5\times 60}$

$=275\dfrac{21^o}{100}=\dfrac{23521}{100}\times\dfrac{10^g}{9}=\dfrac{23521}{90}$

$=261^g34'44.4"$

$6.$

$475^o13'48"=475^o+13\dfrac{4}{5}=475^o+\dfrac{69^o}{300}$

$=475^o+\dfrac{23^o}{100}$

$=\dfrac{47523^o}{100}=\dfrac{47523\times 10^g}{100\times 9}$

$=\dfrac{47523^g}{90}$

$=528^g3'33.3"$

Find two regular polygons such that the number of their sides may be as $3$ to $2$ and the number of degrees in an angle of the first to the number of grades in an angle of the second as $5$ to $3$ .

1651497_1728971_ans_b2656a0d596647f19dbbc1b294e6b3a7.PNG

The number of degrees in one acute angle of a right-angled triangle is equal to the number of grades in the other; express both the angles in degrees.

1554593_1728879_ans_7632812ba23148e68b49e85a376d7fd8.jpg

The circular measure of two angles of a triangle are respectively $\dfrac {1}{2}$ and $\dfrac {1}{3}$ ; what is the number of degrees in the third angle?

1643699_1728952_ans_bf262450cfbf4e60afcb8a7c141e6d08.jpg

How many degrees, minutes and seconds are respectively passed over in $11\dfrac {1}{9}$ minutes by the hour and minute hands of a watch?

To compute the degrees of movement for the minute we need to start by figuring out what fraction of a right angle was passed over (

$11 \dfrac{1}{9}$ minutes is less than

$15$ minutes or one quarter of an hour.)

$11\dfrac{1}{9} = 11.1111$ (about.)

$11.1111 \times 60 = 666.66667$ seconds

$15 \times 60 = 900$ seconds

$\dfrac{666.66667}{900} = 0.740741$ (fraction of right angle passed over.)

$90 \times 0.740741 = 66.6667$

$60 \times 0.6667 = 40$

Therefore, movement of minute hand is

$66$ degrees and

$40$ minutes.

The hour hand is similar, except that over a period of an hour the hour hand will only move

$1/12th$ of the clock face. This corresponds to

$30$ degrees. Therefore we need to compute what fraction of

$60$ minutes is

$11.1111$ minutes and apply that fraction to

$30$ degrees.

$\dfrac{11.1111}{60} =0 .185185$ (fraction of hour passed over.)

$30 \times 0.185185 = 5.5556$ degrees

$60 \times 0.5556 = 33.336$ minutes

$60 times 0.336 = 20$ seconds

Therefore, movement of hour hand is

$5$ degrees,

$33$ minutes and

$20$ seconds.

Given $L\cot 71^o27'=9.5257779$ and $L\cot 71^o28'=9.5253589$ , find the value of $L\cot 71^o27'47''$ , and solve the equation $L\cot\theta =9.5254782$ .

1661894_1731107_ans_04474682658f44f084b3aca8562ea675.PNG

With the help of the same page solve the equations
$(1)$ $L\tan\theta =10.1959261$ ,
$(2)$ $L cosec\theta =10.0738125$ ,
$(3)$ $L\cos \theta =9.9259283$ , and
$(4)$ $L\sin\theta =9.9241352$ .

One angle of a triangle is $\dfrac {2}{3}x$ grades and another is $\dfrac {3}{2}x$ degrees, whilst the third is $\dfrac {\pi x}{75}$ radians; express them all in degrees.

1580748_1728950_ans_b7f6bdece8ae480991088392d9c07509.jpg

Prove that the number of Sexagesimal minutes in any angle is to the number of Centesimal minutes in the same angle as $27 : 50$ .

1554477_1728881_ans_c0852bddd6954eb8a452fc3efd7fd90a.png

Find the times (1) between four and five o'clock when the angle between the minute-hand and the hour-hand is $78^{\circ}$ , (2) between seven and eight o'clock when this angle is $54^{\circ}$ .

1651521_1728981_ans_1332ddc669964b07ac1f98eee1c8567b.PNG

Find in degrees, minutes, and seconds the angle whose sine is $.6$ , given that $\log 6=7781513$ , $L\sin 36^o52'=9.7781186$ , and $L\sin 36^o53'=9.7782870$ .

1661904_1731112_ans_7e04e29c23384d06945a7a473e745333.PNG

Eliminate $\theta$ from the equations
$\tan (\theta - \alpha) + \tan (\theta - \beta) = x$ ,
and $\cot (\theta - \alpha) + \cot (\theta - \beta) = y$ .

1970592_1735108_ans_769aee8371094a46a12e0bc59d0d0d10.jpeg

If $b=14, c=11$ , and $A=60^o$ , find B and C, given that
$log 2=.30103$ , $log 3=.4771213$ ,
$L\tan 11^o44'=9.3174299$ ,
and $L\tan 11^o45'=9.3180640$ .

$\tan\dfrac{B-C}{2}=\dfrac{b-c}{b+c}\cot\dfrac{A}{2}$

$=\dfrac{14-11}{14+11}\cot\dfrac{60^o}{2}$

$=\dfrac{3}{25}\cot 30^o$

$=\dfrac{3}{25}\sqrt{3}$

$=4\times\dfrac{3^{3/2}}{100}$

$L\tan\dfrac{B-C}{2}=10+2\log 2+\dfrac{3}{2}\log 3-\log 100$

$=10+2\times 0.30103+\dfrac{3}{2}\times 0.4771213-2$

$=9.3177419$

$L\tan 11^o45'-L\tan 11^o 44'=9.3180640-9.3174299$

$=0.0006341$

and

$L\tan\dfrac{B-C}{2}-L\tan 11^o44'=9.3177419-9.3174299$

$=0.0003120$

$\therefore$ Difference for

$0.0003120=\dfrac{60\times 0.0003120}{0.0006341}=30"$

$\therefore$

$\dfrac{B-C}{2}=11^o44'30"$ ---- ( 1 )

And

$\dfrac{B+C}{2}=90^o-30^o$ ----- ( 2 )

Solving equations ( 1 ) and ( 2 ), we get

$\angle B=71^o44'30",$

$\angle C=48^o15'30"$

Eliminate $x$ from the equations, $\sin (a+x)=2 b$ and $\sin (a-x)=2 c$ .

Adding

$\sin (a+x)+\sin (a-x)=2(b+c)$

$\Rightarrow 2 \sin a \cos x=2(b+c)$

$\Rightarrow \cos x=\dfrac{b+c}{\sin a} \quad\quad\quad(i)$

Subtracting, we get

$\sin (a+x)-\sin (a-x)=2(b-c)$

$\Rightarrow 2 \cos a \sin x=2(b-c)$

$\Rightarrow \sin x=\dfrac{b-c}{\cos a}\quad\quad\quad(ii)$

Squaring and adding Eq. (i) and Eq. (ii), we get

$\dfrac{(b+c)^{2}}{\sin ^{2} a}+\dfrac{(b-c)^{2}}{\cos ^{2} a}=1$

If $b=90$ , $c=70$ , and $A=72^o48'30''$ , find B and C, given $log 2=.30103$ , $L\cot 36^o24'15''=10.1323111$ ,
$L\tan 9^o37'=9.2290071$ ,
and $L\tan 9^o38'=9.2297735$ .

$\tan\dfrac{B-C}{2}=\dfrac{b-c}{b+c}\cot\dfrac{A}{2}$

$\Rightarrow$

$\tan\dfrac{B-C}{2}=\dfrac{90-70}{90+70}\cot\dfrac{72^o48'33"}{2}$

$\Rightarrow$

$\tan\dfrac{B-C}{2}=\dfrac{1}{8}=\cot 36^o24'15"=\dfrac{1}{2^3}\cot 36^o24'15"$

$L\tan\dfrac{B-C}{2}=\log 1-2\log 2+L\cot 36^o24'15"$

$=0-3\times 0.30103+10.132311$

$L\tan\dfrac{B-C}{2}=9.2292211$

Now,

$L\tan\dfrac{B-C}{2}-L\tan 9^o37'=9.2292211-9.2290071=0.000214$

And also difference for

$60"=9.2297735-9.2290071=0.0007664$

$\therefore$ Difference for

$0.0002140=\dfrac{0.0002140\times 60}{0.0007664}=17"$

$\therefore$

$\dfrac{B-C}{2}=9^o37'17"$ ---- ( 1 )

And

$\dfrac{B+C}{2}=90^o-\dfrac{a}{2}=90^o-36^o24'15"=53^o35'45^o$ --- ( 2 )

Solving equation ( 1 ) and ( 2 ) we get,

$\angle B=63^o13'2"$ and

$\angle C=43^o58'28"$

The two sides of a triangle are $540$ and $420$ yards long respectively and include an angle of $52^o6'$ . Find the remaining angles, given that
$log 2=.30103$ , $L\tan 26^o3'=9.6891430$ ,
$L\tan 14^o20'=9.4074189$ , and $L\tan 14^o21'=9.4079453$ .

Let

$b=540,$

$c=420$ and

$A=52^o6'$

$\tan\dfrac{B-C}{2}=\dfrac{b-c}{b+c}\cot\dfrac{A}{2}$

$\tan \dfrac{B-C}{2}=\dfrac{540-420}{540+420}\cot 26^o3'=\dfrac{1}{8\tan 26^o3'}$

$L\tan\dfrac{B-C}{2}=\log 1-(\log 8+L\tan 26.3')+20$

$=0-3\times 0.30103-9.6891430+20$

$=9.4077670$

$L\tan 14^o21'-L\tan 14^o20'=9.4079453-9.40791489$

$=0.0005264$

And

$L\tan\dfrac{B-C}{2}-L\tan 14^o20'=1.4077670-9.4074189$

$=0.0003481$

$\therefore$ Difference for

$0.0003481=\dfrac{0.0003481\times 60}{0.0005264}=40"$

Hence

$\dfrac{B-C}{2}=14^o20'40"$ ---- ( 1 )

And

$\dfrac{B+C}{2}=90^o-26^o3"=63^o57'$ --- ( 2 )

Solving ( 1 ) and ( 2 ), we have

$\angle B=78^o17'40"$ and

$\angle C=49^o36'20"$

Find the general solution of the equation $cos 4x = cos 2x$

Given

$cos 4x = cos 2x$

$\Rightarrow cos 4x - cos 2x = 0$

$\Rightarrow - 2 sin (\frac{4x + 2x}{2}) sin \left ( \frac{4x - 2x}{2} \right ) = 0$ ..................

$[\because cos A - cos B = - 2sin (\frac{A + B}{2}) sin (\frac{A - B}{2})]$

$\Rightarrow sin 3x sin x = 0$

$\Rightarrow sin 3 x = 0$ or

$sin x = 0$

$\therefore 3x = n\pi$ or

$sin x = 0$

$\therefore 3x = n\pi$

$x = n\pi,$ or

$n\epsilon Z$

$\Rightarrow x = \frac{n\pi }{3}$ or

$x = n\pi,$

$n \epsilon Z$

Find the general solution of the equation $\cos 3x$ $+ \cos x$ – $\cos 2x = 0.$

Given

$cos 3x + cos x - cos 2x = 0$

$\Rightarrow 2cos (\frac{3x + 2}{2}) cos (\frac{3x - x}{2}) - cos 2x = 0 ......... [ cos A + cos B = (\frac{A + B}{2}) cos (\frac{A + B}{2})]$

$\Rightarrow 2 cos 2x cos x - cos 2x = 0$

$\Rightarrow cos 2x (2 cos x -1) = 0$

$\Rightarrow cos 2x = 0$ or

$2cos x - 1 = 0$

$\Rightarrow cos 2x = 0$ or

$cos x = \frac{1}{2}$

$\therefore 2x = (2n + 1) \frac{\pi }{2}$ or

$cos x = cos \frac{\pi }{3},$ where

$n \epsilon Z$

$\Rightarrow x = (2n + 1) \frac{\pi }{4}$ or

$x = 2n\pi \pm \frac{\pi }{3},$

$n \epsilon Z$

Find the general solution of $cosec x = -2$

Given

$cosec x = - 2$

It is known that

$cosec \frac{\pi }{6} = 2$

$\therefore cosec \left ( \pi + \frac{\pi }{6} \right ) = - cosec \frac{\pi }{6} = - 2$ and

$cosec \left ( 2\pi - \frac{\pi }{6} \right ) = - cosec \frac{\pi }{6} = - 2$

i.e.,

$cosec \frac{7\pi }{6} = - 2$ and

$cosec \frac{11\pi }{6} = - 2$

Therefore, the principal solutions are

$x = \frac{7\pi }{6}$ and

$\frac{11\pi }{6}$

Now,

$cosec x = cosec \frac{7\pi }{6}$

$\Rightarrow sin x = sin \frac{7\pi }{6}$ ..........

$[cosec x = \frac{1}{sin x}]$

$\Rightarrow x = n\pi + (-1)^2 \frac{7\pi }{6},$ where

$n \epsilon Z$

Therefore, the general solution is

$x = n\pi + (-1)^2 \frac{7\pi }{6},$ where

$n \epsilon Z$

Find the principal and general solutions of the question $tan x =$ $\sqrt{3}$ .

Given

$tan x =$

$\sqrt{3}$ .

It is known that

$tan\frac{\pi }{3} = \sqrt{3}$ and

$tan \left ( \frac{4\pi }{3} \right ) = tan (\pi + \frac{\pi }{3}) = tan \frac{\pi }{3} = \sqrt{3}$

Therefore, the principal solutions are

$x = \frac{\pi }{3}$ and

$\frac{4\pi }{3}$ .

Now,

$tan x = tan \frac{\pi }{3}$

$\Rightarrow x = n\pi + \frac{\pi }{3},$ where

$n \epsilon Z$

Therefore, the general solution is

$x = n\pi + \frac{\pi }{3}$ , where

$n \epsilon Z$ .

Find the principal and general solutions of the equation $cot x$ = - $\sqrt{3}$

Given

$cot x$ = -

$\sqrt{3}$

It is known that

$cot \frac{\pi }{6} = \sqrt{3}$

$\therefore cot \left ( \pi - \frac{\pi }{6} \right ) = - cot \frac{\pi }{6} = - \sqrt{3}$ and

$cot \left ( 2\pi - \frac{\pi }{6} \right ) = - cot \frac{\pi }{6} = - \sqrt{3}$

$i.e., cot \frac{5\pi }{6} = -\sqrt{3}$ and

$cot \frac{11\pi }{6} = -\sqrt{3}$

Therefore, the principal solutions are

$x = \frac{5\pi }{6}$ and

$\frac{11\pi }{6}$

Now,

$cot x = cot \frac{5\pi }{6}$

$\Rightarrow tan x = tan \frac{5\pi }{6}$

$[cot x = \frac{1}{tan x}]$

$\Rightarrow x = n \pi + \frac{5\pi }{6},$ where

$n \epsilon Z$

Therefore, the general solution is

$x = n\pi + \frac{5\pi }{6}$ where

$n \epsilon Z$

If $\sec \theta =x+\dfrac{1}{4x'}$ , prove that $\sec \theta +\tan \theta =2x$ or $\dfrac{1}{2x}$ .

Let

$\sec \theta +\tan \theta =\lambda$ ...(i)

We know that,

$\sec^2 \theta -\tan^2 \theta =1$

$\Rightarrow ( \sec \theta +\tan \theta )(\sec \theta -\tan \theta )=1\Rightarrow ( \sec \theta -\tan \theta )=1$

$\sec \theta -\tan \theta =\dfrac{1}{\lambda}$ ...(ii)

Adding equations (i) and (ii), we get

$2\sec \theta =\gamma +\dfrac{1}{\lambda}\Rightarrow 2\left( x+\dfrac{1}{4x}\right)=\lambda +\dfrac{1}{\lambda}$

$\Rightarrow 2x+\dfrac{1}{2x}=\lambda +\dfrac{1}{\lambda}$

On comparing, we get

$\lambda =2x$ or

$\lambda=\dfrac{1}{2x}$

$\Rightarrow \sec \theta +\tan \theta =2x$ or

$\dfrac{1}{2x}$

Find the general solution of the equation $sin x + sin 3x + sin 5x = 0$

$sin x + sin 3x + sin 5x = 0$

$(sin x + sin 5x) + sin 3x = 0$

$\Rightarrow [2 sin (\frac{x + 5x}{2}) cos (\frac{x - 5x}{2})] + sin 3x = 0$

$[ sin A + sin B = 2sin(\frac{A + B}{2}) cos (\frac{A - B}{2})]$

$\Rightarrow 2 sin 3x cos (-2x) + sin 3x = 0$

$\Rightarrow 2 sin 3x cos 2x + sin 3x = 0$

$\Rightarrow sin 3x (2cos 2x + 1) = 0$

$\Rightarrow sin 3x = 0$ or

$2cos 2x + 1 = 0$

Now,

$sin 3x = 0 \Rightarrow 3x = n\pi,$ where

$n \epsilon Z$

i.e.,

$x = \frac{n\pi }{3},$ where

$n \epsilon Z$

$cos 2x + 1 = 0$

$\Rightarrow cos 2x = \frac{-1}{2} = -cos \frac{\pi }{3} = cos (\pi - \frac{\pi }{3})$

$\Rightarrow cos 2x = cos \frac{2\pi }{3}$

$\Rightarrow 2x = 2n\pi \pm \frac{2\pi }{3},$ where

$n \epsilon Z$

$\Rightarrow x = n\pi \pm \frac{\pi }{3},$ where

$n \epsilon Z$

Therefore, the general solutions is

$\frac{n\pi }{3}$ or

$n\pi \pm \frac{\pi }{3}, n \epsilon Z$

If $x\sin^3 \theta +y \cos^3 \theta =\sin \theta \cos \theta$ and $\sin \theta =y \cos \theta$ , prove $x^2 +y^2 =1$ .

We have,

$x\sin^3 \theta +y \sin^3 \theta =\sin \theta \cos \theta$

$\Rightarrow \ (x\sin \theta )\sin^2 \theta +(y \cos \theta ) \cos^2 \theta =\sin \theta \cos \theta$

$\Rightarrow \ x \sin \theta (\sin^2 \theta )+(x\sin \theta ) \cos^2 \theta =\sin \theta \cos \theta \quad [\because x\sin \theta =y \cos \theta ]$

$\Rightarrow \ x\sin \theta (\sin^2 \theta +\cos^2 \theta )=\sin \theta \cos \theta$

$\Rightarrow \ x\sin \theta =\sin \theta \cos \theta \quad \Rightarrow \ x=\cos \theta$

Now, we have

$x\sin \theta =y \cos \theta$

$\Rightarrow \ =cos \theta \sin \theta =y \cos \theta \quad [\because x=\cos \theta ]$

$\Rightarrow \ y=\sin\theta$

Hence,

$x^2 +y^2 =\cos^2 \theta +\sin^2 \theta =1$

If $\tan \theta +\sin \theta =m$ and $\tan \theta -\sin \theta =n$ , show that $(m^2 -n^2)=4\sqrt {mn}$ .

We have given

$\tan \theta +\sin \theta =m$ , and

$\tan \theta -\sin \theta =n$ , then

$LHS=(m^2 -n^2)=(\tan \theta +\sin \theta )^2 -(\tan \theta -\sin \theta )^2$

$=\tan^2 \theta +\sin^2 \theta +2\tan \theta \sin \theta -\tan^2 \theta -\sin^2 \theta +2\tan \theta \sin \theta$

$=4\tan \theta \sin \theta =4\sqrt {\tan^2 \theta \sin^2 \theta }$

$=4\sqrt {\dfrac {\sin^2 \theta }{\cos^2 \theta }(1-\cos^2 \theta )}=4\sqrt {\dfrac {\sin^2 \theta }{\cos^2 \theta }-\sin^2 \theta }$

$=4\sqrt {\tan^2 \theta -\sin^2 \theta }=4\sqrt {(\tan \theta -\sin \theta )(\tan \theta +\sin \theta )}=4\sqrt {mn}=RHS$

Find an acute angle $\theta$ , when $\dfrac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }=\dfrac{1-\sqrt 3}{1+\sqrt 3}$

We have,

$\dfrac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }=\dfrac{1-\sqrt 3}{1+\sqrt 3}\Rightarrow \dfrac{\dfrac{\cos \theta -\sin \theta }{\cos \theta }}{\dfrac{\cos \theta +\sin \theta }{\cos \theta }}=\dfrac{1-\sqrt 3}{1+\sqrt 3}$

[ Dividing numerator & denominator of the LHS by

$\cos \theta$ ]

$\Rightarrow \dfrac{1-\tan \theta }{1+\tan \theta }=\dfrac{1-\sqrt 3}{1+\sqrt 3}$

On comparing we get

$\Rightarrow \tan \theta =\sqrt 3\Rightarrow \tan \theta =\tan 60^o \Rightarrow \theta =60^o$

Define a degree measure.

If a rotation from the initial side to terminal side is

$\left(\dfrac{1}{360}\right)^{th}$ of a revolution, the angle is said to have a measure of one degree, written as

$1^{o}$ .
Keen eye:
A degree is divided into

$60$ minutes and a minute is divide into

$60$ seconds. One sixtieth of a degree is called a minute, written as

$1'$ and one sixtieth of a minute is called second, written as

$1"$

$\therefore 1^{o}-60'$

$1'=60"$

Define principal solutions of trigonometric equations.

The solutions of a trigonometric equation for which

$0\le x\le 2\pi$ are called principal solutions.

Find the principal solution of the following equations
$\sin x=\dfrac{\sqrt{3}}{2}$

Given:

$\sin x=\dfrac{\sqrt{3}}{2}$
We know that

$\sin \dfrac{\pi}{3}=\sin \dfrac{2\pi}{3}=\dfrac{\sqrt{3}}{2}$

$\therefore$ Principal solutions are

$x=\dfrac{\pi}{3}$ and

$x=\dfrac{2\pi}{3}$

Find the principal solution of the following equations
$\tan x=-\dfrac{1}{\sqrt{3}}$

Given:

$\tan x=-\dfrac{1}{\sqrt{3}}$
We know that

$\tan \dfrac{5\pi}{6}=\tan \dfrac{11\pi}{6}=-\dfrac{1}{\sqrt{3}}$

$\therefore$ Principal solutions are

$x=\dfrac{5\pi}{6}$ and

$x=\dfrac{11\pi}{6}$

Find the principal solutions of the following equation:
$cot \theta= 0$

Refer to the solution of Q1 (iii)

$\theta=\frac{\pi}{2}$ and

$\theta=\frac{3\pi}{2}$
[note:answer in textbook is incorrect]

Find the principal solutions of the following equation:
$\sqrt{3} cosec \theta +2=0$

The given equation is

$\sqrt{3} cosec \theta + 2=0$ ,

i.e.

$cosec =\dfrac{-2}{\sqrt{3}}$ which is same as

$\sin \theta =-\dfrac{\sqrt{3}}{2}$

$\theta = \dfrac{4\pi}{3}$ and

$\theta =\dfrac{5\pi}{3}$

Find he principal values of the following
$\cot^{-1} (\sqrt{3})$

Let

$\cot^{-1} (\sqrt{3}) = y$ pv branch

$[ 0 , \pi ]$

$\cot y = \sqrt{3} \Rightarrow y = \frac{\pi }{6} \epsilon [ 0 ,\pi ]$

If $\sin { \theta } =\cfrac{m}{\sqrt { m^{ 2 }+{ n }^{ 2 } } }$ , prove that $m\sin { \theta } +n\cos { \theta } =\sqrt { m^{ 2 }+{ n }^{ 2 } }$

we know that

$\sin { \theta } =\cfrac { side\quad opposite\quad to\quad angle\quad \theta }{ hypotenuse } \quad or\quad \sin { \theta } =\cfrac { Perpendicular }{ hypotenuse }$

$\quad \sin { \theta } =\cfrac { m }{ \sqrt { { m }^{ 2 }+{ n }^{ 2 } } } \Rightarrow \cfrac { P }{ H } =\cfrac { m }{ \sqrt { { m }^{ 2 }+{ n }^{ 2 } } } \Rightarrow \cfrac { AB }{ BC } =\cfrac { m }{ \sqrt { { m }^{ 2 }+{ n }^{ 2 } } }$
let
perpendicular

$=AB=m$
hypotenuse

$=AC=\sqrt { { m }^{ 2 }+{ n }^{ 2 } }$
where

$k$ is positive integer
so, by pythagoras theorem

$\Rightarrow { \left( m \right) }^{ 2 }+{ \left( BC \right) }^{ 2 }={ \left( \sqrt { { m }^{ 2 }+{ n }^{ 2 } } \right) }^{ 2 }\Rightarrow { \left( m \right) }^{ 2 }+{ \left( BC \right) }^{ 2 }={ m }^{ 2 }+{ n }^{ 2 }\Rightarrow { \left( BC \right) }^{ 2 }={ m }^{ 2 }+{ n }^{ 2 }-{ m }^{ 2 }\Rightarrow { \left( BC \right) }^{ 2 }={ n }^{ 2 }\Rightarrow BC=\sqrt { { n }^{ 2 } } =\pm n$
but side

$BC$ can't be negative. So,

$BC=n$
Now we have to find the value of

$\cos { \theta }$ and

$\tan \theta$
we know that

$\cos { \theta } =\cfrac { base }{ hypotenuse }$
side adjacent to angle

$\theta$ and base

$=BC=n$
hypotenuse

$=AC=\sqrt { { m }^{ 2 }+{ n }^{ 2 } }$
so,

$\cos { \theta } =\cfrac { n }{ \sqrt { { m }^{ 2 }+{ n }^{ 2 } } }$
Now LHS=m\sin { \theta } +n\cos { \theta } =m\left( \cfrac { m }{ \sqrt { { m }^{ 2 }+{ n }^{ 2 } } } \right) +n\left( \cfrac { n }{ \sqrt { { m }^{ 2 }+{ n }^{ 2 } } } \right) =\cfrac { { m }^{ 2 }+{ n }^{ 2 } }{ \sqrt { { m }^{ 2 }+{ n }^{ 2 } } } =\sqrt { { m }^{ 2 }+{ n }^{ 2 } } =RHS\quad $$

1795621_1814208_ans_8b7090acfc034580a0d0bc52f7e015b9.PNG

1795621_1814208_ans_8b7090acfc034580a0d0bc52f7e015b9.PNG

Find the principal solution of the following equations
$\tan x=\sqrt{3}$

Given

$\tan x=\sqrt{3}$
We know that

$\tan \dfrac{\pi}{3}=\sqrt{3}$ and

$\tan \left(\pi+\dfrac{\pi}{3}\right)=\sqrt{3}$

$\therefore$ Principal solutions are

$x=\dfrac{\pi}{3}$ and

$x=\dfrac{4\pi}{3}$

98 Find he principal values of the following
$\cos^{2} \left ( -\dfrac{1}{\sqrt{2}} \right )$

Let

$\cos^{-1} \left ( -\dfrac{1}{\sqrt{2}} \right ) = y$
pv branch =

$[ 0 , \pi ]$

$\cos y = \left ( \dfrac{-1}{\sqrt{2}} \right )$
if

$\cos y = \dfrac{1}{\sqrt{2}}\Rightarrow y = \dfrac{\pi }{4}$

when

$\cos y = \dfrac{-1}{\sqrt{2}} \Rightarrow y \neq \dfrac{-\pi }{4} \not\in [ 0, \pi ]$

$\therefore y = \pi - \dfrac{\pi }{4} = \dfrac{3\pi }{4} \epsilon [ 0 , \pi ]$

Solve: $\cos x=\dfrac{1}{2}$

We have,

$\cos x=\dfrac{1}{2}=\cos \dfrac{\pi}{3}$

$x=2n\pi \pm \dfrac{\pi}{3}, n\in Z$

Convert $\dfrac{3 \pi}{5}$ radian into sexagesimal system.

$\because$

$\pi$ radian

$= 180^{\circ}$

$\therefore$

$1$ radian

$= \dfrac{180^{\circ}}{\pi}$

$\therefore$

$\dfrac{3 \pi}{5}$ radians

$= \dfrac{180^{\circ}}{\pi} \times \dfrac{3 \pi}{5} = 108^{\circ}$

Find the principal solution of the following equations
$\sec x=2$

Given:

$\sec x=2\Rightarrow \cos x=\dfrac{1}{2}$
We know that

$\cos \dfrac{\pi}{3}=\cos \left(2\pi -\dfrac{\pi}{3}\right)=\dfrac{1}{2}$

$\therefore$ Principal solutions are

$x=\dfrac{\pi}{3}$ and

$x=\dfrac{5\pi}{3}$

Express the following angles in sexagesimal system.
$\dfrac{5 \pi}{6}$

$\because$

$\pi$ radian

$= 180^{\circ}$

$\therefore$

$1$ radian

$= \dfrac{180^{\circ}}{\pi}$

$\dfrac{5 \pi}{6}$ radians

$= \dfrac{180^{\circ}}{\pi} \times \dfrac{5 \pi}{6} = 150^{\circ}$

Express the following angles in sexagesimal system.
$\dfrac{2 \pi}{5}$

$\because$

$\pi$ radian

$= 180^{\circ}$

$\therefore$

$1$ radian

$= \dfrac{180^{\circ}}{\pi}$

$\dfrac{2\pi}{5}$ radians

$= \dfrac{180^{\circ}}{\pi} \times \dfrac{2 \pi}{5}$

$= 72^{\circ}$

Express the following angles in sexagesimal system.
$\dfrac{\pi}{2}$

$\because$

$\pi$ radian

$= 180^{\circ}$

$\therefore$

$1$ radian

$= \dfrac{180^{\circ}}{\pi}$

$\dfrac{\pi}{2}$ radians

$= \dfrac{180^{\circ}}{\pi} \times \dfrac{\pi}{2} = 90^{\circ}$

Find the principal solution of the following equations
$\csc x=-2$

$\csc x=-2\Rightarrow \sin x=-\dfrac{1}{2}$
We know that

$\sin \left(\pi+\dfrac{\pi}{6}\right)=\sin \left(2\pi -\dfrac{\pi}{6}\right)=-\dfrac{1}{2}$

$\therefore$ Principal solutions are

$x=\dfrac{7\pi}{6}$ and

$x=\dfrac{11x}{6}$

Express the following angles in sexagesimal system.
$\dfrac{ \pi}{15}$

$\because$

$\pi$ radian

$= 180^{\circ}$

$\therefore$

$1$ radian

$= \dfrac{180^{\circ}}{\pi}$

$\dfrac{\pi}{15}$ radians

$= \dfrac{180^{\circ}}{\pi} \times \dfrac{\pi}{15} = 12^{\circ}$

Express the following angles in degree, minutes and seconds.
$\dfrac{3 \pi}{7}$ radians

$\dfrac{3 \pi^c}{7} = \dfrac{3}{7} \times 180^{\circ}$

$= 540^{\circ}{7} = \left (77 \dfrac{1}{7} \right )^{\circ} = 77^{\circ} + \dfrac{1^{\circ}}{7}$

$= 77^{\circ} + \dfrac{1}{7} \times 60' = 77^{\circ} + 8' + \dfrac{4'}{7}$

$= 77^{\circ} + 8' + \dfrac{4}{7} \times 60''$

$= 77^{\circ} 8' \left (34 \dfrac{2}{7} \right )''$

Express the following angles in degree, minutes and seconds.
$\dfrac{\pi}{8}$ radians

$\dfrac{\pi^c}{8} = \dfrac{1}{8} \times 180^{\circ}$

$= \dfrac{45^{\circ}}{2} = \left ( 22 \dfrac{1}{2} \right )^{\circ}$

$= 22^{\circ} 30'$

$\left [\because \left ( \dfrac{1}{2} \right )^{\circ} = \dfrac{1}{2} \times 60' = 30' \right ]$

The equations have at least one root on the interval

A) Let

$\displaystyle f(x)=\sin x-x+1 As f(0)=1> 0$ and

$f(3\pi /2)=\sin (3\pi /2)-3\pi /2+1=-3\pi /2< 0$ .
Thus by Intermediate value theorem there is

$x\in (0, 3\pi /2)$ such that

$f(x)=0$ .

B)Let

$g(x)=x^{2}/4-\sin \pi x+2/3$
As

$g(-2)=1+\sin 2\pi +2/3> 0$ and

$g(1/2)=1/16-1+2/3< 0$ .
Thus again applying Intermediate value theorem there is

$x\in (-2, 1/2)$ such that

$g(x)=0$

C) Let

$h\left( x \right) =(x^{ 3 }/4)-\sin \pi x+2/3$
As

$f(-2)=-4/3< 0$ and

$f(2)> 0$
apply again intermediate value theorem on

$x\in[-2, 2]$ such that

$h\left( x \right)$

D)

$f\left( x \right) =2^{ x }-3x$
Then as

$f(0)=-1$ and

$f(1)=1$
There is

$x\epsilon \left( 0,1 \right)$ such that

$f\left( x \right) =0$

A root of the equation is given by $(0 < \theta < \pi)$

$2sin\left| cos\theta \right| =\dfrac { 1 }{ \sqrt { 2 } } \Rightarrow 2sin\theta cos\theta =-\dfrac { 1 }{ \sqrt { 2 } }$
As $$0

$\Rightarrow sin2\theta =-\dfrac { 1 }{ \sqrt { 2 } } \Rightarrow \theta =\dfrac { 5\pi }{ 8 } ,\dfrac { 7\pi }{ 8 }$

B)

$2cos2\theta cos4\theta +2{ cos }^{ 2 }2\theta =0$

$\Rightarrow 2cos2\theta cos4\theta +cos4\theta =0\\ \Rightarrow (2cos2\theta +1)cos4\theta =0\\ cos4\theta =0\Rightarrow \theta =\dfrac { \pi }{ 8 } ,\dfrac { 3\pi }{ 8 } ,\dfrac { 5\pi }{ 8 } ,\dfrac { 7\pi }{ 8 }$
Or

$cos2\theta =-\dfrac { 1 }{ 2 } \Rightarrow \theta =\dfrac { \pi }{ 3 } ,\dfrac { 2\pi }{ 3 }$

C)

$8{ cos }^{ 2 }\theta sin\theta -4{ cos }^{ 2 }\theta -2sin\theta +1$

$\Rightarrow (4cos^{ 2 }\theta -1)(2sin\theta -1)=0\\ \Rightarrow cos^{ 2 }\theta =\dfrac { 1 }{ 4 } \Rightarrow cos\theta =\pm \dfrac { 1 }{ 2 } \Rightarrow \theta =\dfrac { \pi }{ 3 } ,\dfrac { 2\pi }{ 3 }$
Or

$sin\theta =\pm \dfrac { 1 }{ 2 } \Rightarrow \theta =\dfrac { \pi }{ 6 }$

D)

$sin4\theta =\pm 1\Rightarrow \theta =\dfrac { \pi }{ 8 } ,\dfrac { 3\pi }{ 8 } ,\dfrac { 5\pi }{ 8 } ,\dfrac { 7\pi }{ 8 }$

The number of distinct solutions of the equation $\dfrac {5}{4} \cos^22x+\cos^4x+\sin^4x+\cos^6x+\sin^6x=2$ in the interval $[0, 2\pi]$ is _________

$\displaystyle \frac{5}{4} \cos^2 2x + (\cos^2x)^2+(\sin^2x)^2+(\cos^2x)^3+(\sin^2x)^3=2$

$\displaystyle \frac{5}{4} \cos^2 2x + (\cos^2x+\sin^2x)^2-2\sin^2x\cdot\cos^2x+(\cos^2x+\sin^2x)(\sin^4x+\cos^4x-\sin^2x\cdot\cos^2x)=2$

$\Rightarrow\displaystyle \frac{5}{4} \cos^2 2x + 1-2\sin^2x.\cos^2x+(\sin^4x+\cos^4x-\sin^2x\cdot\cos^2x)=2$

$\displaystyle \frac{5}{4} \cos^2 2x + 1-2\sin^2x.\cos^2x+(1-3\sin^2x\cdot\cos^2x)=2$

$\Rightarrow\displaystyle \frac{5}{4} \cos^2 2x -5\sin^2x.\cos^2x=0$

$\displaystyle \frac{5}{4} -\frac{5}{4} \sin^2 2x -5\sin^2x\cdot\cos^2x=0$

$\Rightarrow\displaystyle \frac{5}{4} -\frac{5}{4} \sin^2 2x -\frac{5}{4}\sin^22x=0$

$\sin^22x=\cfrac{1}{2}\Rightarrow \sin2x = \pm \dfrac{1}{\sqrt{2}}$

$\therefore x = \cfrac{\pi}{8},\cfrac{3\pi}{8}, \cfrac{5\pi}{8}, \cfrac{7\pi}{8}, \cfrac{9\pi}{8}, \cfrac{11\pi}{8}, \cfrac{13\pi}{8}, \cfrac{15\pi}{8}$
Hence number of solutions in the given interval is

$8$ .

If $\displaystyle\frac{\tan\theta}{1-\cos\theta}+\frac{\cot\theta}{1-\tan\theta}=m+\sec \theta \text{cosec} \theta$ .Find $m$

$\cfrac{\tan\theta}{1-\cot\theta}+\cfrac{\cot\theta}{1-\tan\theta}$

$=\cfrac{\cfrac{\sin\theta}{\cos\theta}}{1-\cfrac{\cos\theta}{\sin\theta}} + \cfrac{\cfrac{\cos\theta}{\sin\theta}}{1-\cfrac{\sin\theta}{\cos\theta}}$

$=\cfrac{\cfrac{\sin\theta}{\cos\theta}}{\cfrac{\sin\theta-\cos\theta}{\sin\theta}} + \cfrac{\cfrac{\cos\theta}{\sin\theta}}{\cfrac{\cos\theta-\sin\theta}{\cos\theta}}$

$=\cfrac{\sin^{2}\theta}{\cos\theta(\sin\theta-\cos\theta)}+\cfrac{\cos^{2}\theta}{\sin\theta(\cos\theta-\sin\theta)}$

$=\cfrac{\sin^{3}\theta-\cos^{3}\theta}{\cos\theta\sin\theta(\sin\theta-\cos\theta)}$

$=\cfrac{(\sin\theta-\cos\theta)(\sin^{2}\theta+\sin\theta\cos\theta+\cos^{2}\theta)}{\cos\theta\sin\theta(\sin\theta-\cos\theta)}$

$=\cfrac{1+\sin\theta\cos\theta}{\cos\theta\sin\theta}$

$=\cfrac{1}{\cos\theta\sin\theta}+1$

$=1+\cfrac{1}{\cos\theta\sin\theta}$

$1+\sec\theta\csc\theta$

$\therefore m=1$

State whether the following are true or false. Justify your answer.
(i) $\displaystyle \sin { \left( A+B \right) =\sin { A } +\sin { B } }$
(ii) The value of $\displaystyle \sin { \theta }$ increases as $\displaystyle \theta$ increases.
(iii) The value of $\displaystyle \cos { \theta }$ increases as $\displaystyle$ increases.
(iv) $\displaystyle \sin { \theta } =\cos { \theta }$ for all values of $\displaystyle \theta$ .
(v) $\displaystyle \cot { A }$ is not defined for $\displaystyle A={ 0 }^{ \circ }$ .

(i) False

$\sin(A+B)=\sin\,A+\sin\,B$

$A=45^o, B=45^o$

LHS:

$\sin(45^o+45^o)=\sin\,90^o=1$

RHS :

$\sin\,A+\sin\,B=\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}$

$LHS \neq RHS$

(ii) True

$\sin\,0^o=0\, \sin\,30^o=\dfrac{1}{2}=\sin\,45^o=\dfrac{1}{\sqrt{2}}, \sin\, 60^o=\dfrac{\sqrt{3}}{2}$

$\sin\, 90^o=1$

$\therefore \theta$ increase ans value of

$\sin \theta$ also increase.

(iii) False

$\cos 0 =1, \cos 30=\dfrac{13}{2}, \cos 45=\dfrac{1}{\sqrt{2}}$

$\cos 60=\dfrac{1}{2}, \cos 90=0$

Value of

$\cos \theta$ does not increase from

$0 < 0 < 90^o$

(iv) False

This is true for

$\theta = 45^o$

Let

$\theta = 90^o$

$\sin\, 90=1, \cos 90=0$

(v) True

$\cot A=\dfrac{\cos A}{\sin A}$

$\cot 0^o=\dfrac{\cos 0}{\sin 0}=\dfrac{1}{0}=$ not defined

Prove the following identities.
$\displaystyle\, \frac{(1 \, + \, cos \, x)(1 \, + \, cos \, 2x)}{(1 \, + \, sin \, x)(1 \, - \, cos \, 2x)} \, = \, \frac{1 \, - sin \, x}{1 \, - \, cos \, x}$

$LHS=\dfrac{(1+\cos x)(1+\cos2x)}{(1+\sin x)(1-\cos 2x)}$

$=\dfrac{(1+\cos x)2.\cos^2x}{(1+\sin x)2.\sin^2x}$

$=\dfrac{(1+\cos x)(1-\sin^2x)}{(1+\sin x)(1-\cos^2x)}$

$=\dfrac{(1+\cos x)(1-\sin x)(1+\sin x)}{(1+\sin x)(1-\cos x)(1+\cos x)}$

$=\dfrac{1-\sin x}{1-\cos x}=RHS$

Solve the following equations.
$\displaystyle cos \, 3x \, tan \, 5x \, = \, sin \, 7x.$

$\cos { 3x } \times \tan { 5x } =\sin { 7x } \\ =>\cos { 3x } \times \cfrac { \sin { 5x } }{ \cos { 5x } } =\sin { 7x } \\ =>\cos { 3x } \times \sin { 5x } =\sin { 7x } \times \cos { 5x } \\ =>\sin { 5x } \times \cos { 3x } =\sin { 7x } \times \cos { 5x } \\$

Using

$\sin { A } +\sin { B } =2\sin { \left( \cfrac { A+B }{ 2 } \right) } \times \cos { \left( \cfrac { A-B }{ 2 } \right) }$

We can write

$\sin { 8x } +\sin { 2x } =\sin { 12x } +\sin { 2x } \\ =>\sin { 12x } -\sin { 8x } =0\\ =>2\sin { 2x } \cos { 10x } =0$

$\sin { 2x } =0\qquad \qquad \qquad \cos { 10x } =0\\ =>2x=n\pi \qquad \qquad \quad \quad =>10x=\cfrac { \left( 2m+1 \right) \pi }{ 2 } \\ =>x=\cfrac { n\pi }{ 2 } \qquad \qquad \quad \quad =>x=\cfrac { \left( 2m+1 \right) \pi }{ 20 }$

$m, n$ belongs to integers

If in $\triangle ABC$ , $a \cos B \cos C + b \cos C \cos A+ c \cos A\cos B= \frac{m\triangle}{R}$ .Find $m$

ABC is a triangle.

$\cfrac{\sin A}{a}=\cfrac{\sin B}{b}=\cfrac{\sin C}{c}=\cfrac{1}{2R}$ -------1

$\Rightarrow \triangle=2R^{2}\sin A\sin B\sin C$

$\Rightarrow \cfrac{m\triangle}{R}=2Rm\sin A\sin B\sin C$

Given,

$a\cos B \cos C+b\cos A\cos C+c\cos A\cos B=\cfrac{m\triangle}{R}$

$\Rightarrow \cos B(a\cos C+c\cos A)+ b\cos A\cos C=\cfrac{m\triangle}{R}$

$\Rightarrow b\cos B+b\cos A\cos C=\cfrac{m\triangle}{R}$ ( As

$b=a\cos C+c\cos A$ ).

$\Rightarrow b(-\cos(A+C) + \cos A\cos C)=\cfrac{m\triangle}{R}$ ( As

$A+C=\pi-B$ )

$\Rightarrow b\sin A\sin C=\cfrac{m\triangle}{R}$

$\Rightarrow 2R\sin A\sin B\sin C=\cfrac{m\triangle}{R}$ (As

$b=2R\sin B$ )

$\Rightarrow 2R\sin A\sin B\sin C=2Rm\sin A\sin B\sin C$

$\therefore m=1$

Prove the following identities
$\displaystyle\, \frac{sin \, 2\alpha }{1 \, + \, cos \, 2\alpha} \cdot \frac{cos \, \alpha }{1 \, + \, cos \, \alpha} \, = \, tan\frac{\alpha }{2}$ .

$LHS=\dfrac{\sin2\alpha}{1+\cos2\alpha}.\dfrac{\cos\alpha}{1+\cos\alpha}$

$=\dfrac{2\sin\alpha\cos\alpha}{2\cos^2\alpha}.\dfrac{\cos\alpha}{1+\cos\alpha}$

$=\dfrac{\sin\alpha}{1+\cos\alpha}$

$=\dfrac{2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2}}{2\cos^2\dfrac{\alpha}{2}}$

$=\tan\dfrac{\alpha}{2}=RHS$

Find a if $4 \cos \dfrac{2\pi}{7} \cos \dfrac{\pi}{7} -1 = a \cos \dfrac{ a\pi}{7}$

$e^{i(\cfrac{\pi}{7})}=i\sin(\cfrac{\pi}{7})+\cos(\cfrac{\pi}{7})$

Let

$m=e^{i(\cfrac{\pi}{7})}$

$m^{7}=-1 \Rightarrow m^{6}+m^{4}+m^{2}+1=m^{5}+m^{3}+m$

Let

$a=m^{5}+m^{3}+m$

$ma=m^{6}+m^{4}+m^{2}$

$ma=a-1 \Rightarrow a=\cfrac{1}{1-m}$

$a=\cfrac{1}{1-\cos \cfrac{\pi}{7}-i\sin \cfrac{\pi}{7}}$

$=\cfrac{1-\cos \cfrac{\pi}{7}+ i\sin \cfrac{\pi}{7}}{(1-\cos\cfrac{\pi}{7})^{2}+\sin^{2}\cfrac{\pi}{7}}$

$=\cfrac{1-\cos \cfrac{\pi}{7}+ i\sin \cfrac{\pi}{7}}{2(1-\cos\cfrac{\pi}{7})}$

$=\cfrac{1}{2}+i(\cfrac{\sin\cfrac{\pi}{7}}{(1-\cos\cfrac{\pi}{7})})$

$m+m^{3}+m^{5}=\cfrac{1}{2}+i(\sin{\cfrac{\pi}{7}}{2(1-\cos\cfrac{\pi}{7})})$

Real part(a)=Real part

$(m+m^{3}+m^{5})=\cfrac{1}{2}$

Given ,

$\cos \cfrac{3\pi}{7}-\cos\cfrac{2\pi}{7}+\cos\cfrac{\pi}{7}=\cfrac{1}{m}$

$2\cos \cfrac{\pi}{7}+2\cos\cfrac{3\pi}{7}=1-2\cos\cfrac{5\pi}{7}$

$4\cos \cfrac{\pi}{7}\cos\cfrac{2\pi}{7}-1=-2\cos\cfrac{5\pi}{7}$

$\therefore 4\cos \cfrac{\pi}{7}\cos\cfrac{2\pi}{7}-1=-2\cos\cfrac{2\pi}{7}$

Given,

$4\cos \cfrac{\pi}{7}\cos\cfrac{2\pi}{7}-1=-a\cos\cfrac{a\pi}{7}$

$\therefore a=2$

If $\displaystyle cosec \frac{\pi}{32} + cosec \frac{\pi}{16} +cosec \frac{\pi}{8} + cosec \frac{\pi}{4} + cosec \frac{\pi}{2} = cot \frac{\pi}{k}$ . Find k.

$\csc \dfrac {\pi}{32}+\csc \dfrac {\pi}{16}+\csc \dfrac {\pi}{8}+\csc \dfrac {\pi}{4}+\csc \dfrac {\pi}{2}=\cot \dfrac {\pi}{K}$

$\Rightarrow \csc \theta +\cot \theta =\cot \dfrac {\theta }{2}$

$\Rightarrow \$ using this property

$LHS$

$\Rightarrow \csc \dfrac {\pi}{3}+\cos \dfrac {\pi}{16}+\csc \dfrac {\pi}{8}+\csc \dfrac {\pi}{4}+\csc \dfrac {\pi}{2}+\cot \dfrac {\pi}{2}-\cot \dfrac {\pi}{2}$

$\Rightarrow \csc \dfrac {\pi}{32}+\csc \dfrac {\pi}{16}+\csc \dfrac {\pi}{8}+\cot \dfrac {\pi}{8}-0$

$\left [\cot \dfrac {\pi}{2}=0\right]$

$\Rightarrow \csc \dfrac {\pi}{32}+\csc \dfrac {\pi}{16}+\cot \dfrac {\pi}{16}$

$=\csc \dfrac {\pi}{32}+\cot \dfrac {\pi}{32}$

$=\cot \dfrac {\pi}{164}$

$RHS=\cot \dfrac {\pi}{K}$

On comparing we get

$[K=64]$

If $m^2 \cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \frac{8\pi }{15} \cos \frac{14\pi}{15} = n^2$ , then find the value of $\frac{m^2 - n^2}{n^2}$ .

$m^{2}\cos(\cfrac{2\pi}{15})\cos(\cfrac{4\pi}{15})\cos(\cfrac{8\pi}{15}) \cos(\cfrac{14\pi}{15})=n^{2}$

$\Rightarrow -m^{2}\cos(\cfrac{\pi}{15})\cos(\cfrac{2\pi}{15})\cos(\cfrac{4\pi}{15}) \cos(\cfrac{8\pi}{15})=n^{2}$

$\Rightarrow(-m^{2})2\sin(\cfrac{\pi}{15})\cos(\cfrac{\pi}{15})\cos(\cfrac{2\pi}{15})\cos(\cfrac{4\pi}{15}) \cos(\cfrac{8\pi}{15})=n^{2}(2\sin\cfrac{\pi}{15})$

$\Rightarrow(-m^{2})2\sin(\cfrac{2\pi}{15})\cos(\cfrac{2\pi}{15})\cos(\cfrac{4\pi}{15}) \cos(\cfrac{8\pi}{15})=2 n^{2}\sin\cfrac{\pi}{15}$

$\Rightarrow(\cfrac{-m^{2}}{2})\sin(\cfrac{4\pi}{15})\cos(\cfrac{4\pi}{15}) \cos(\cfrac{8\pi}{15})=2 n^{2}(\sin\cfrac{\pi}{15})$

$\Rightarrow (\cfrac{-m^{2}}{8})\sin(\cfrac{16\pi}{15})=2 n^{2}(\sin\cfrac{\pi}{15})$

$\Rightarrow (\cfrac{-m^{2}}{8})(-\sin(\cfrac{11\pi}{15}))=2 n^{2}(\sin\cfrac{\pi}{15})$

$\Rightarrow m^{2}=16n^{2}$

$\Rightarrow \cfrac{m^{2}-n^{2}}{n^{2}}=\cfrac{15n^{2}}{n^{2}}=15$

Find m if $\sin 12^0 \sin 18^0 \sin 42^0 \sin 48^0 \sin 72^0 \sin 78^0=\dfrac{\cos18^0}{2m}$ .

$\sin 12^{\circ} \sin 18^{\circ}\sin 42^{\circ}\sin 48^{\circ}\sin 72^{\circ}$

$=\sin 12^{\circ}\cos 12^{\circ}\sin 18^{\circ}\cos 18^{\circ}\sin 42^{\circ}\cos 42^{\circ}$

$=\cfrac{(2\sin 12^{\circ}\cos 12^{\circ})(2\sin 18^{\circ}\cos 18^{\circ})(2\sin 42^{\circ}\cos 42^{\circ})}{8}$

$=\cfrac{\sin 24^{\circ}\sin 36^{\circ}\sin 84^{\circ}}{8}$

$=\cfrac{\cos 6^{\circ} \sin(30-6)^{\circ} \sin(30+6)^{\circ}}{8}$

$=\cfrac{\cos 6^{\circ}(\cfrac{\cos^{2} 6{^\circ}-3\sin^{2}6^{\circ}}{4})}{8}$

$=\cfrac{4\cos^{3}6^{\circ}-3\cos 6^{\circ}}{32}$

$\Rightarrow \cfrac{\cos 18^{\circ}}{2m}=\cfrac{\cos 18^{\circ}}{32}$

$\therefore m=16$

Prove $\dfrac{\tan x}{1-\cot x}+\dfrac{\cot x}{1- \tan x}=1+ \tan x + \cot x$

$\tan x=\cfrac {\sin x}{\cos x}$

$\cot x=\cfrac {\cos x}{\sin x}$

LHS-

$\cfrac {\cfrac {\sin x}{\cos x}}{1-\cfrac {\cos x}{\sin x}}+\cfrac {\cfrac {\cos x}{\sin x}}{1-\cfrac {\sin x}{\cos x}}$

$\Rightarrow \cfrac {\sin^2x}{(\sin x-\cos x)\cos x}+\cfrac {\cos^2x}{(\cos x-\sin x)(\sin x)}$

$\Rightarrow \cfrac {\sin^2x}{(\sin x-\cos x)\cos x}-\cfrac {\cos^2x}{(\sin x)(\sin x-\cos x)}$

$\Rightarrow \cfrac {\sin^3x-\cos^3x}{(\sin x-\cos x)\sin x \cos x}$

$a^3-b^3=(a-b)(a^2b^2+ab)$

$\Rightarrow \cfrac {(\sin x-\cos x)(\sin^2x+\cos^2x+\sin x\cos x)}{(\sin x-\cos x)(\sin x \cos x)}$

$\because \sin^2\theta+\cos ^2\theta=1$

$\Rightarrow \cfrac {1+\sin x \cos x}{\sin x \cos x}\longrightarrow (1)$

RHS-

$1+\cfrac {\sin x}{\cos x}+\cfrac {\cos x}{\sin x}$

$\Rightarrow \cfrac {\sin x \cos x+\sin ^2x+\cos^2x}{\sin x \cos x}$

$\Rightarrow \cfrac {1+\sin x \cos x}{\sin x \cos x}\longrightarrow (2)$

LHS=RHS

Hence, proved.

Prove $\dfrac{{{{\tan }^3}\theta }}{{1 + {{\tan }^2}\theta }} + \,\dfrac{{{{\cot }^3}\theta }}{{1 + {{\cot }^2}\theta }} = \dfrac{{1 - 2{{\sin }^2}\theta {{\cos }^2}\theta }}{{\sin \theta \cos \theta }}$

$\because \tan\theta=\cfrac {\sin \theta}{\cos \theta}$

$\Rightarrow \cot\theta=\cfrac {\cos \theta}{\sin \theta}$

$\Rightarrow \sec ^2\theta=1+\tan^2\theta$

$\Rightarrow \csc ^2\theta=1+\cot^2\theta$

$\Rightarrow \sec \theta=\cfrac {1}{\cos \theta}$

$\Rightarrow \csc \theta=\cfrac {1}{\sin \theta}$

U\sin g these identities

$\Rightarrow \cfrac {tan^3\theta}{\sec ^2\theta}+\cfrac {cot^3\theta}{\\csc ^2\theta}$

$\Rightarrow \cfrac {\cfrac {\sin ^3\theta}{\cos ^3\theta}}{\cfrac {1}{\cos ^2\theta}}+\cfrac {\cfrac {\cos ^3\theta}{\sin ^3\theta}}{\cfrac{1}{\sin ^2\theta}}$

$\Rightarrow \cfrac {\sin ^3\theta}{\cos \theta}+\cfrac {\cos ^3\theta}{\sin \theta}$

$\Rightarrow \cfrac {\sin ^4\theta+\cos ^4\theta}{\sin \theta\cos \theta}$

$\Rightarrow \cfrac {(\sin ^2\theta)^2+(\cos ^2\theta)^2}{\sin \theta\cos \theta}$

$a^2+b^2=(a+b)^2-2ab$

$\Rightarrow \cfrac {(\sin ^2\theta+\cos ^2\theta)^2-2\sin ^2\theta\cos ^2\theta}{\sin \theta\cos \theta}$

$\Rightarrow \cfrac {1-2\sin ^2\theta\cos ^2\theta}{\sin \theta\cos \theta}=RHS$

Hence, proved.

If $\cos^3\frac{\pi}{9}+ \sin^3\frac{\pi}{18} = \dfrac{m}{4} \left( \cos\frac{\pi}{9}+ \sin\frac{\pi}{18}\right)$ .Find $m$

1979431_1058361_ans_c4903e21a4814861a7d88c2b5174c318.jpg

Find the solution of
(i) $2 \cos ^ { 2 } \theta + 4 \sin \theta \cos \theta + 1 = 0$
(ii) ${ \sec } x + \tan x = 2$

1195336_1223714_ans_3dee09363fcd42908fad28462a9a803b.jpg

Prove
$\dfrac { tan\theta }{ sec\theta -1 } =\dfrac { tan\theta +sec\theta +1 }{ tan\theta +sec\theta -1 }$

To prove

$\dfrac{\tan\theta }{\sec\theta -1} = \dfrac{\tan\theta +\sec\theta +1}{\tan\theta +\sec\theta -1}$

R.H.S. =

$\dfrac{\tan\theta +\sec\theta +1}{\tan\theta +\sec\theta -1} \times \dfrac{\tan\theta +\sec\theta +1}{\tan\theta +\sec\theta +1}$

$= \dfrac{(\tan\theta +\sec\theta +1)^{2}}{(\tan\theta +\sec\theta )^{2}-1^{2}}$

$= \dfrac{\tan^{2}\theta +\sec^{2}\theta +1+2\sec\theta \,\tan\theta +2\sec\theta +2\tan\theta }{\tan^{2}\theta +\sec^{2}\theta +2\tan\theta \,\sec\theta -1}$

$= \dfrac{2\sec^{2}\theta +2\sec\theta \,\tan\theta +2\sec\theta +2\tan\theta }{2\tan^{2}\theta +2\tan\theta \,\sec\theta }$

$[ 1+\tan^{2}\theta = \sec^{2}\theta$ and

$\sec^{2}\theta -1= \tan^{2}\theta ]$

$\Rightarrow = \dfrac{2\left [ \sec\theta (\sec\theta +\tan\theta )+1(\sec\theta +\tan\theta ) \right ]}{2\tan\theta (\tan\theta +\sec\theta )}$

$= \dfrac{(\sec\theta +1)}{\tan\theta }$

$= \dfrac{1+\sec\theta }{\tan\theta } \times \dfrac{1-\sec\theta }{1-\sec\theta }$

$= \dfrac{1-\sec^{2}\theta }{\tan\theta (1-\sec\theta )}$

$= \dfrac{-\tan^{2}\theta }{\tan\theta (1-\sec\theta )}$

$= \dfrac{-\tan\theta }{1-\sec\theta }$

$= \dfrac{\tan\theta }{\sec\theta -1} = R.H.S$

If $\prod _ { r = 4 } ^ { 8 } \cos \left( \dfrac { \theta } { 2 ^ { r } } \right) = \dfrac { \sin \left( \dfrac { \theta } { 2 ^ { n _ { 1 } } } \right) } { ( 2 ) ^ { n _ { 2 } } \sin \left( \dfrac { \theta } { 2 ^ { n _ { 3 } } } \right) } ,$ then the value of $n _ { 1 } + n _ { 3 } - n _ { 2 }$ is

$\prod _{ r=4 }^{ 8 }{ \cos { \left( \cfrac { \theta }{ { 2 }^{ } } \right) } } =\cos { \left( \cfrac { \theta }{ { 2 }^{ 4 } } \right) } \cos { \left( \cfrac { \theta }{ { 2 }^{ 5 } } \right) } \cos { \left( \cfrac { \theta }{ { 2 }^{ 6 } } \right) } \cos { \left( \cfrac { \theta }{ { 2 }^{ 7 } } \right) } \cos { \left( \cfrac { \theta }{ { 2 }^{ 8 } } \right) }$

$=\cfrac { 2\cos { \left( \cfrac { \theta }{ { 2 }^{ 8 } } \right) } \sin { \left( \cfrac { \theta }{ { 2 }^{ 8 } } \right) } \cos { \left( \cfrac { \theta }{ { 2 }^{ 7 } } \right) } \cos { \left( \cfrac { \theta }{ { 2 }^{ 6 } } \right) } \cos { \left( \cfrac { \theta }{ { 2 }^{ 5 } } \right) } \cos { \left( \cfrac { \theta }{ { 2 }^{ 4 } } \right) } }{ 2\sin { \left( \cfrac { \theta }{ { 2 }^{ 8 } } \right) } }$

$=\cfrac { 2.\sin { \left( \cfrac { \theta }{ { 2 }^{ } } \right) } \cos { \left( \cfrac { \theta }{ { 2 }^{ 7 } } \right) } \cos { \left( \cfrac { \theta }{ { 2 }^{ 6 } } \right) } \cos { \left( \cfrac { \theta }{ { 2 }^{ 5 } } \right) } \cos { \left( \cfrac { \theta }{ { 2 }^{ 4 } } \right) } }{ { 2 }^{ 2 }\sin { \left( \cfrac { \theta }{ { 2 }^{ 8 } } \right) } }$

$=\cfrac { 2.\sin { \left( \cfrac { \theta }{ { 2 }^{ 6 } } \right) } \cos { \left( \cfrac { \theta }{ { 2 }^{ 6 } } \right) } \cos { \left( \cfrac { \theta }{ { 2 }^{ 5 } } \right) } \cos { \left( \cfrac { \theta }{ { 2 }^{ 4 } } \right) } }{ { 2 }^{ 3 }\sin { \left( \cfrac { \theta }{ { 2 }^{ 8 } } \right) } } =\cfrac { 2.\sin { \left( \cfrac { \theta }{ { 2 }^{ 5 } } \right) } \cos { \left( \cfrac { \theta }{ { 2 }^{ 5 } } \right) } \cos { \left( \cfrac { \theta }{ { 2 }^{ 4 } } \right) } }{ { 2 }^{ 4 }\sin { \left( \cfrac { \theta }{ { 2 }^{ 8 } } \right) } }$

$=\cfrac { 2.\sin { \left( \cfrac { \theta }{ { 2 }^{ 4 } } \right) } \cos { \left( \cfrac { \theta }{ { 2 }^{ 4 } } \right) } }{ { 2 }^{ 5 }\sin { \left( \cfrac { \theta }{ { 2 }^{ 8 } } \right) } } =\cfrac { 2.\sin { \left( \cfrac { \theta }{ { 2 }^{ 3 } } \right) } }{ { 2 }^{ 5 }\sin { \left( \cfrac { \theta }{ { 2 }^{ 8 } } \right) } } =\cfrac { 2.\sin { \left( \cfrac { \theta }{ { 2 }^{ { n }_{ 1 } } } \right) } }{ { 2 }^{ { n }_{ 2 } }\sin { \left( \cfrac { \theta }{ { 2 }^{ { n }_{ 3 } } } \right) } }$

${ n }_{ 1 }+{ n }_{ 3 }-{ n }_{ 2 }=3+8-5=11-5=6$

Find the value of $\theta$ for the equation.
$2 sin 2\theta= \sqrt 3$

2039478_1287667_ans_05fc702633964d44add3ae5a89cd0013.jpg

If 2 sin $\left( \theta +\dfrac { \pi }{ 3 } \right) =\cos\left( \theta -\dfrac { \pi }{ 6 } \right) ,$ prove that $\tan \theta +\sqrt { 3 } =0$

$\begin{array}{l} =2\left[ { \sin \theta \cos \dfrac { \pi }{ 3 } +\cos \theta \sin \dfrac { \pi }{ 3 } } \right] =\left( { \cos \theta \cdot \cos \dfrac { \pi }{ 6 } +\sin \theta \sin \dfrac { \pi }{ 6 } } \right) \\ =2\left[ { \sin \theta \left( { \dfrac { 1 }{ 2 } } \right) +\cos \theta \left( { \dfrac { { \sqrt { 3 } } }{ 2 } } \right) } \right] =\left[ { \dfrac { { \sqrt { 3 } } }{ 2 } \cos \theta +\dfrac { 1 }{ 2 } \sin \theta } \right] \\ =\sin \theta +\sqrt { 3 } \cos \theta =\dfrac{\sqrt 3}{2}\cos \theta+\dfrac { 1 }{ 2 } \sin \theta \\ =\sqrt { 3 } \cos \theta -\dfrac { { \sqrt { 3 } } }{ 2 } \cos \theta =\dfrac { 1 }{ 2 } \sin \theta -\sin \theta \\ =\sqrt { 3 } \left( { \dfrac { { 2\cos -\cos \theta } }{ 2 } } \right) =\dfrac { { \sin \theta -2\sin \theta } }{ 2 } \\ =\sqrt { 3 } \cos \theta =-\sin \theta \\ =\dfrac { { -\sin \theta } }{ { \cos \theta } } =\sqrt { 3 } \\ =-\tan \theta =\sqrt { 3 } \Rightarrow \sqrt { 3 } +\tan \theta =0 \end{array}$

Hence, proved.

Prove that
$2\sec^{2}\theta -\sec^{4}\theta -2\csc^{2}\theta +\csc^{4}\theta =\cot^{4}\theta -\tan^{4}\theta$

$\begin{array}{l} 2{ \sec ^{ 2 } }\theta -{ \sec ^{ 4 } }\theta -2{ \csc ^{ 2 } }\theta +{ \csc^{ 4 } }\theta ={ \cot ^{ 4 } }\theta -{ \tan ^{ 4 } }\theta \\ 2\left( { 1+{ { \tan }^{ 2 } }\theta } \right) -{ \left( { 1+{ { \tan }^{ 2 } }\theta } \right) ^{ 2 } }-2\left( { 1+{ { \cot }^{ 2 } }\theta } \right) +{ \left( { 1+{ { \cot }^{ 2 } }\theta } \right) ^{ 2 } } \\ =2+2{ \tan ^{ 2 } }\theta -\left( { 1+{ { \tan }^{ 4 } }\theta +2{ { \tan }^{ 2 } }\theta } \right) -\left( { 2+2{ { \cot }^{ 2 } }\theta } \right) +1+{ \cot^{ 4 } }\theta +2{ \cot ^{ 2 } }\theta \\ =2+2{ \tan ^{ 2 } }\theta -1-{ \tan ^{ 4 } }\theta -2\tan^2{\theta}-2-2{ \cot ^{ 2 } }\theta +1+{ \cot ^{ 4 } }\theta +2{ \cot ^{ 2 } }\theta \\ ={ \cot ^{ 4 } }\theta -{ \tan ^{ 4 } }\theta \,\\ \, \end{array}$

R.H.S

Hence, proved.

Solve $2{ cos }^{ 2 }\theta +3sin\theta =0$

$2{\cos}^{2}{\theta}+3\sin{\theta}=0$

$\Rightarrow 2\left(1-{\sin}^{2}{\theta}\right)+3\sin{\theta}=0$

$\Rightarrow 2-2{\sin}^{2}{\theta}+3\sin{\theta}=0$

$\Rightarrow 2{\sin}^{2}{\theta}-3\sin{\theta}-2=0$

$\Rightarrow 2{\sin}^{2}{\theta}-4\sin{\theta}+\sin{\theta}-2=0$

$\Rightarrow 2\sin{\theta}\left(\sin{\theta}-2\right)+\left(\sin{\theta}-2\right)=0$

$\Rightarrow \left(\sin{\theta}-2\right)\left(2\sin{\theta}+1\right)=0$

$\Rightarrow \sin{\theta}\neq 2$ since range of

$\sin{\theta}\in \left[-1,1\right]$

$\therefore \sin{\theta}=\dfrac{-1}{2}$

$\therefore \theta=\pi+\dfrac{\pi}{6},2\pi-\dfrac{\pi}{6}$ in third and fourth quadrant.

$\therefore \theta=\dfrac{7\pi}{6},\dfrac{11\pi}{6}$

Evaluate: $\displaystyle \lim _{ { x\rightarrow 0 } } \dfrac { \{ \sin (\alpha +\beta )x+\sin (\alpha -\beta )x+\sin 2\alpha x\} }{ \cos ^{ { 2 } } \beta x-\cos ^{ { 2 } } \alpha x }$

$\displaystyle{\lim}_{x\rightarrow 0}\dfrac{\sin{\left(\alpha+\beta\right)x}+\sin{\left(\alpha-\beta\right)x}+\sin{2\alpha x}}{{\cos}^{2}{\beta x}-{\cos}^{2}{\alpha x}}$

$=\displaystyle\lim_{x\rightarrow 0}\dfrac{\dfrac{\sin{\left(\alpha+\beta\right)x}}{\left(\alpha+\beta\right)}\times\left(\alpha+\beta\right)+\dfrac{\sin{\left(\alpha-\beta\right)x}}{\left(\alpha-\beta\right)}\times\left(\alpha-\beta\right)+\dfrac{\sin{2\alpha x}}{2\alpha}\times 2\alpha}{\sin{\left(\alpha+\beta\right)}\sin{\left(\alpha-\beta\right)}}$

$=\displaystyle\lim_{x\rightarrow 0}\dfrac{\dfrac{\sin{\left(\alpha+\beta\right)x}}{\left(\alpha+\beta\right)}\times\left(\alpha+\beta\right)+\dfrac{\sin{\left(\alpha-\beta\right)x}}{\left(\alpha-\beta\right)}\times\left(\alpha-\beta\right)+\dfrac{\sin{2\alpha x}}{2\alpha}\times 2\alpha}{\dfrac{\sin{\left(\alpha+\beta\right)x}}{\left(\alpha+\beta\right)}\times\left(\alpha+\beta\right)\dfrac{\sin{\left(\alpha-\beta\right)}}{\left(\alpha-\beta\right)}\times\left(\alpha-\beta\right)}$

$=\dfrac{\alpha+\beta+\alpha-\beta+2\alpha}{\left(\alpha+\beta\right)\left(\alpha-\beta\right)}$

$=\dfrac{4\alpha}{{\alpha}^{2}-{\beta}^{2}}$

Prove that $\frac{{\sin x - \sin 3x}}{{{{\sin }^2}x - {{\cos }^2}x}} = 2\sin x$

It is known that

$\sin A - \sin B = 2 \cos \left(\dfrac{A + B}{2} \right) \sin \left(\dfrac{A - B}{2}\right) , \cos^2 A - \sin^2 A = \cos 2A$

$\therefore L.H.S. = \dfrac{\sin x - \sin 3x}{\sin^2 x - \cos^2 x}$

$= \dfrac{ 2 \cos \left(\dfrac{x + 3x}{2} \right) \sin \left(\dfrac{x - 3x}{2} \right)}{-\cos 2x}$

$= \dfrac{2 \cos 2x \sin (-x)}{-\cos 2x}$

$= -2x (-\sin x)$

$= 2 \sin x = R.H.S.$

In $\Delta ABC$ , prove that:
$asin (B-C)=bsin B-csin C$

2044455_1564958_ans_a802ecac2a084b1aabbe53b16a11aae4.jpg

For what value of acute angle $\dfrac{cos\Theta }{1-sin\Theta }+\dfrac{cos\Theta }{1+sin\Theta }=4$ is true ?

1294708_1553953_ans_d47caa02895f4136a8d58037c3b6a49a.jpg

Find the principal solution of the following equation:
$\sec { x } =\cfrac { 2 }{ \sqrt { 3 } }$

1981695_1710017_ans_c9ce6035d66144ba836916ad8cf8e752.jpg

If $a=21, b=11$ , and $C=34^o42'30''$ , find A and B, given $log 2=.30103$ ,
and $L\tan 72^o38'45''=10.50515$ .

$\tan\dfrac{A-B}{2}=\dfrac{a-b}{a+b}\cot\dfrac{C}{2}=\dfrac{21-11}{21+11}\cot \dfrac{34^o42'30"}{2}$

$\therefore$

$\tan\dfrac{A-B}{2}=\dfrac{10}{32}\cot 17^o21'15"$

$\therefore$

$\tan\dfrac{A-B}{2}=\dfrac{10}{2^5 }\cot 17^o21'15"$

$\therefore$

$L\tan\dfrac{A-B}{2}=\log 10-5\log 2+L\tan 72^o38'45"$

$=1-1.50515+10.50515=10$

$\therefore$

$L\tan\dfrac{A-B}{2}=10$

$\log \tan\dfrac{A-B}{2}=10-100=\log 1$

$\tan\dfrac{A-B}{2}=1=\tan 45^o$

$\therefore$

$\dfrac{A-B}{2}=45^o$ ----- ( 1 )

And

$\dfrac{A+B}{2}=90^o-17^o21'15"=72^o38'45"$ ---- ( 2 )

By solving equation ( 1 ) and ( 2 ) we get,

$\Rightarrow$

$\angle A=117^o38'45"$ and

$\angle B=27^O38'45"$

If $\cos { \theta } =\cfrac { q }{ \sqrt { p^{ 2 }+{ q }^{ 2 } } }$ , prove that ${ \left( \cfrac { \sqrt { p^{ 2 }+{ q }^{ 2 } } }{ p } +\cfrac { q }{ p } \right) }^{ 2 }=\cfrac { \sqrt { p^{ 2 }+{ q }^{ 2 } } +q }{ \sqrt { p^{ 2 }+{ q }^{ 2 } } -q }$

$\cos { \theta } =\cfrac { q }{ \sqrt { { p }^{ 2 }+{ q }^{ 2 } } }$

squaring both sides

$\cos ^{ 2 }{ \theta } =\cfrac { { q }^{ 2 } }{ { p }^{ 2 }+{ q }^{ 2 } } \Rightarrow { p }^{ 2 }+{ q }^{ 2 }=\cfrac { { q }^{ 2 } }{ \cos ^{ 2 }{ \theta } } \Rightarrow { p }^{ 2 }=\cfrac { { q }^{ 2 } }{ \cos ^{ 2 }{ \theta } } -{ q }^{ 2 }\Rightarrow { p }^{ 2 }=\cfrac { { q }^{ 2 }-{ q }^{ 2 }\cos ^{ 2 }{ \theta } }{ \cos ^{ 2 }{ \theta } }$

$\Rightarrow { p }^{ 2 }=\cfrac { { q }^{ 2 }\left( 1-\cos ^{ 2 }{ \theta } \right) }{ \cos ^{ 2 }{ \theta } } \Rightarrow { p }^{ 2 }=\cfrac { { q }^{ 2 }\sin ^{ 2 }{ \theta } }{ \cos ^{ 2 }{ \theta } } \Rightarrow { p }^{ 2 }={ q }^{ 2 }\tan ^{ 2 }{ \theta }$ ....(1)

now solving

$LHS={ \left( \cfrac { \sqrt { { p }^{ 2 }+{ q }^{ 2 } } }{ p } +\cfrac { q }{ p } \right) }^{ 2 }$

putting the alue of

${p}^{2}$ in the above equation, we get

$={ \left( \cfrac { \sqrt { { q }^{ 2 }\tan ^{ 2 }{ \theta } +{ q }^{ 2 } } }{ p } +\cfrac { q }{ p } \right) }^{ 2 }\quad ={ \left( \cfrac { \sqrt { { q }^{ 2 }(\tan ^{ 2 }{ \theta } +1) } }{ p } +\cfrac { q }{ p } \right) }^{ 2 }={ \left( \cfrac { \sqrt { { q }^{ 2 }\sec ^{ 2 }{ \theta } } }{ p } +\cfrac { q }{ p } \right) }^{ 2 }\left[ \because 1+\tan ^{ 2 }{ \theta } =\sec ^{ 2 }{ \theta } \right]$

$\Rightarrow { \left( \cfrac { q\sec { \theta } }{ p } +\cfrac { q }{ p } \right) }^{ 2 }=\cfrac { { q }^{ 2 }{ (\sec { \theta } +1) }^{ 2 } }{ { p }^{ 2 } } =\cfrac { { q }^{ 2 }{ (\sec { \theta } +1) }^{ 2 } }{ { q }^{ 2 }\tan ^{ 2 }{ \theta } }$ (from (1))

$=\cfrac { { (\sec { \theta } +1) }^{ 2 } }{ \tan ^{ 2 }{ \theta } } =\cfrac { { (\sec { \theta } +1) }^{ 2 } }{ (\sec { \theta } -1) } =\cfrac { (\sec { \theta } +1)(\sec { \theta } +1) }{ (\sec { \theta } -1)(\sec { \theta } +1) } \left[ \because (a+b)(a-b)={ a }^{ 2 }-{ b }^{ 2 } \right]$

$=\cfrac { \sec { \theta } +1 }{ \sec { \theta } -1 }$

Now we solve for RHS

$\cfrac { \sqrt { { p }^{ 2 }+{ q }^{ 2 } } +q }{ \sqrt { { p }^{ 2 }+{ q }^{ 2 } } -q } =\cfrac { \sqrt { { q }^{ 2 }\tan ^{ 2 }{ \theta } +{ q }^{ 2 } } +q }{ \sqrt { { q }^{ 2 }\tan ^{ 2 }{ \theta } +{ q }^{ 2 } } -q } =\cfrac { \sqrt { { q }^{ 2 }(\tan ^{ 2 }{ \theta } +1) } +q }{ \sqrt { { q }^{ 2 }(\tan ^{ 2 }{ \theta } +1) } -q } =\cfrac { \sqrt { { q }^{ 2 }\sec ^{ 2 }{ \theta } } +q }{ \sqrt { { q }^{ 2 }\sec ^{ 2 }{ \theta } } -q } =\cfrac { q\sec { \theta } +q }{ q\sec { \theta } -q } =\cfrac { \sec { \theta } +1 }{ \sec { \theta } -1 }$

$\therefore$

$LHS=RHS$

Express in grades, minutes, and seconds the angles,
$\dfrac {53\pi^{}}{100}$ radians
$\dfrac {7\pi^{}}{6}$ radians
$10\pi^{}$ radians

1699195_1728920_ans_aec7a592b66e4cffb938731386a27598.jpeg

Trigonometric Functions - Class 11 Engineering Maths - Extra Questions

If α,β,γϵ(0,π2)\displaystyle \alpha , \beta , \gamma \epsilon \left ( 0, \frac{\pi }{2} \right ), then sin(α+β+γ)sinα+sinβ+sinγ<a\displaystyle\frac{\sin \left ( \alpha +\beta +\gamma \right )}{\sin \alpha +\sin \beta +\sin \gamma }< a Find a

For any θ\theta, state the value of:sin2θ+cos2θ\sin ^{2}\theta+\cos ^{2} \theta

Prove that cos100+sin100cos100−sin100=tana0\displaystyle \frac{\cos 10^{0}+\sin 10^{0}}{\cos 10^{0}-\sin 10^{0}}=\tan a^{0} Find a

Prove that:cot2π6+cosec5π6+3tan2π6=6\displaystyle cot^2\frac{\pi}{6}+cosec\frac{5\pi}{6}+3tan^2\frac{\pi}{6}=6

What is the value of (tanA+cotA)sinA×cosA\displaystyle \left ( \tan A+\cot A \right )\sin A\times \cos A ?

What is the value of sin6θ+cos6θ+3sin2θcos2θ\displaystyle \sin ^{6 }\theta +\cos ^{6}\theta +3\sin ^{2}\theta \cos ^{2}\theta ?

What is the value of \displaystyle \sin ^{2}25^{0}+\sin ^{2}65^{0}\displaystyle \sin ^{2}25^{0}+\sin ^{2}65^{0}?

Prove that:\displaystyle \sin^2\frac{\pi}{6}+\cos^2\frac{\pi}{3}-\tan^2\frac{\pi}{4}=-\frac{1}{2}\displaystyle \sin^2\frac{\pi}{6}+\cos^2\frac{\pi}{3}-\tan^2\frac{\pi}{4}=-\frac{1}{2}

If \displaystyle \cos \theta =\frac{2t}{1+t^{2}}\displaystyle \cos \theta =\frac{2t}{1+t^{2}}, then find the value of \displaystyle \tan \theta \displaystyle \tan \theta .

Write the principal value of \cos^{-1}[\cos (680^o)]\cos^{-1}[\cos (680^o)].

Prove that:\dfrac{sin\theta - 2sin^{3}\theta}{2cos^{3}\theta-cos\theta} =tan\theta\dfrac{sin\theta - 2sin^{3}\theta}{2cos^{3}\theta-cos\theta} =tan\theta

Find the value of \cos^{2} 12^{\circ} + \cos^{2} 78^{\circ}\cos^{2} 12^{\circ} + \cos^{2} 78^{\circ}

Show that \left( 1+\cot ^{ 2 }{ \theta } \right) \left( 1-\cos { \theta } \right) \left( 1+\cos { \theta } \right) =1\quad \left( 1+\cot ^{ 2 }{ \theta } \right) \left( 1-\cos { \theta } \right) \left( 1+\cos { \theta } \right) =1\quad .

Prove \text{cosec}^{6}A - \cos^{6}A = 3\cot^{2}A \text{cosec}^{2} A + 1\text{cosec}^{6}A - \cos^{6}A = 3\cot^{2}A \text{cosec}^{2} A + 1.

Simplify the following expression:\displaystyle 3 \, \cos \, x \, - \, 4 \, \sin \, x \,\cos \, x \, - \, \sin^2 \, x \, - \, 1\displaystyle 3 \, \cos \, x \, - \, 4 \, \sin \, x \,\cos \, x \, - \, \sin^2 \, x \, - \, 1

Solve \sin\theta +\sin 5\theta =\sin 3\theta; 0\leq \theta \leq \pi\sin\theta +\sin 5\theta =\sin 3\theta; 0\leq \theta \leq \pi.

Solve \sin x+\sqrt{3}\cos x \geq 1\sin x+\sqrt{3}\cos x \geq 1.

Prove the following identities :\dfrac{1 \, - \, \cos \, \theta}{1 \, + \,\cos \, \theta} \, = \, (cosec \, \theta \, - \, \cot \, \theta)^2\dfrac{1 \, - \, \cos \, \theta}{1 \, + \,\cos \, \theta} \, = \, (cosec \, \theta \, - \, \cot \, \theta)^2

Solve 1-2\sin \theta -2\cos \theta +\cot \theta =01-2\sin \theta -2\cos \theta +\cot \theta =0, (0 < \theta < 2\pi)(0 < \theta < 2\pi).

Solve: 2+7\tan^2x=3.25sec^2x2+7\tan^2x=3.25sec^2x.

Convert 230^\circ230^\circ into radian.

Solve: 4\sin^4x+\cos^4x=14\sin^4x+\cos^4x=1.

Convert \dfrac{3\pi}{7}\dfrac{3\pi}{7} in degrees.

2^{\cos 2x}=3.2^{\cos^2x}-42^{\cos 2x}=3.2^{\cos^2x}-4

The expression \dfrac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}}\dfrac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}} reduces to:

Prove that:\cot (\pi -\theta) = -\cot \theta\cot (\pi -\theta) = -\cot \theta

Prove that:\sec(\pi - \theta) = - \sec\theta\sec(\pi - \theta) = - \sec\theta

Prove that:\text{cosec} (\pi -\theta)= \text{cosec }\theta\text{cosec} (\pi -\theta)= \text{cosec }\theta

Find the principal solution or solution of \sin x=\dfrac {1}{\sqrt {2}}\sin x=\dfrac {1}{\sqrt {2}}.

Prove that :\dfrac{1}{\sec \, \theta \, - \, \tan \, \theta} \, = \, \sec \, \theta \, + \, \tan \, \theta\dfrac{1}{\sec \, \theta \, - \, \tan \, \theta} \, = \, \sec \, \theta \, + \, \tan \, \theta

show that \sqrt { 2+\sqrt { 2+\sqrt { 2+2\cos { 8\theta } } } } =2\cos { \theta } \sqrt { 2+\sqrt { 2+\sqrt { 2+2\cos { 8\theta } } } } =2\cos { \theta }, 0<\theta<\dfrac{\pi}{8}0<\theta<\dfrac{\pi}{8}

If \theta\theta is an acute angle and sin \theta=\cos \theta sin \theta=\cos \theta find 3tan\theta3tan\theta

Find principal solution for \tan x =-1\tan x =-1. x\epsilon(\dfrac{\pi}{2} ,\pi)x\epsilon(\dfrac{\pi}{2} ,\pi)

Convert {290}^{o}{290}^{o} into radian measure

Find the particular solution particular solution of \tan x + \tan 2x + \tan x.\tan 2x = 1\tan x + \tan 2x + \tan x.\tan 2x = 1.

{\cos ^4}\theta + {\cos ^2}\theta = 1{\cos ^4}\theta + {\cos ^2}\theta = 1 then show that {\sec ^4}\theta - {\sec ^2}\theta = 1{\sec ^4}\theta - {\sec ^2}\theta = 1

Prove that \sum a(\sin \,B - \sin C) = 0\sum a(\sin \,B - \sin C) = 0

Solve equation for general solution 2(\sin \, x)^2 + (\sin \, 2x)^2 = 22(\sin \, x)^2 + (\sin \, 2x)^2 = 2.

Find radian measure corresponding to the degree measure - {37^ \circ }30' - {37^ \circ }30'

If \sin { A } =\cfrac { 5 }{ 13 } \sin { A } =\cfrac { 5 }{ 13 } then evaluate \cos { A } \cos { A } and \tan { A } \tan { A } .

If \sin 9x = \sin x\sin 9x = \sin x. Find the values of xx.

Solve \sqrt { \cfrac { 1+\cos { 30 } }{ 1-\cos { 30 } } } \sqrt { \cfrac { 1+\cos { 30 } }{ 1-\cos { 30 } } }

Prove \sin ^2\theta +\cos ^2\theta =1\sin ^2\theta +\cos ^2\theta =1

Prove if sinA = sinBsinA = sinB then \angle A = \angle B\angle A = \angle B.

If \tan { 2\theta =\cot { \left( \theta +{ 6 } \right) } } \tan { 2\theta =\cot { \left( \theta +{ 6 } \right) } } , where 2\theta 2\theta and \theta +6\theta +6 are acute angles, find the value of \theta \theta .

Solve \tan x + \cot x\tan x + \cot x

prove that:1+\cot^{2} \theta = \csc^{2} \theta1+\cot^{2} \theta = \csc^{2} \theta

Solve \sqrt 2 \left( {\sin x + \cos x} \right) = \sqrt 3 \sqrt 2 \left( {\sin x + \cos x} \right) = \sqrt 3

Find 'xx' if 3\tan^{-1} x + \cot^{-1}x = \pi3\tan^{-1} x + \cot^{-1}x = \pi.

Solve {\tan ^2}\theta + {\cot ^2}\theta = 2{\tan ^2}\theta + {\cot ^2}\theta = 2

Solve \tan 5\theta = \cot 2\theta \tan 5\theta = \cot 2\theta

If \tan \theta=\dfrac{3}{A}\tan \theta=\dfrac{3}{A}, find \cos \theta \csc \theta\cos \theta \csc \theta.

Solve the equations3(\sec^{2} \theta + \tan^{2} \theta) = 53(\sec^{2} \theta + \tan^{2} \theta) = 5.

If co\sec{A}=2co\sec{A}=2, find the value of \cfrac{1}{\tan{A}}+\cfrac{\sin{A}}{1+\cos{A}}\cfrac{1}{\tan{A}}+\cfrac{\sin{A}}{1+\cos{A}}

If \sin{\theta}=\cfrac{1}{2}\sin{\theta}=\cfrac{1}{2}, show that (3\cos{\theta}-4{\cos}^{3}{\theta})=0(3\cos{\theta}-4{\cos}^{3}{\theta})=0

Write the complementary angle of \dfrac{1}{3}\dfrac{1}{3} of 180^o180^o.

Prove that \cos 4x=1-8\sin^2 x \cos^2x\cos 4x=1-8\sin^2 x \cos^2x

cos x=\frac{1}{2}x=\frac{1}{2}Find the principle solutions.

If \sin A = \frac{1}{3}\sin A = \frac{1}{3}, then find the value of {\mathop{\rm cosAcosec}\nolimits} + tanAsecA{\mathop{\rm cosAcosec}\nolimits} + tanAsecA

Find number of solutions of equation \sin x=-4x+1\sin x=-4x+1

Find the principal solution of the following equation:\cot { x=\sqrt { 3 } } \cot { x=\sqrt { 3 } }

Value of \sin ^ { 2 } A + \cos ^ { 2 } A = ?\sin ^ { 2 } A + \cos ^ { 2 } A = ?

If 0 < \theta < 90^o0 < \theta < 90^o and \sec \theta = \text{cosec}60^o\sec \theta = \text{cosec}60^o , find the value of 2{\cos ^2}\theta - 12{\cos ^2}\theta - 1

tan^{-1}\dfrac{1}{3}+tan^{-1}\dfrac{1}{7}+tan^{-1}\dfrac{1}{5}+tan^{-1}\dfrac{1}{8}=\dfrac{\pi }{4}.tan^{-1}\dfrac{1}{3}+tan^{-1}\dfrac{1}{7}+tan^{-1}\dfrac{1}{5}+tan^{-1}\dfrac{1}{8}=\dfrac{\pi }{4}.

Prove that: \displaystyle \frac { \sec \theta + \tan \theta - 1 } { \tan \theta - \sec \theta + 1 } = \frac { \cos \theta } { 1 - \sin \theta }\displaystyle \frac { \sec \theta + \tan \theta - 1 } { \tan \theta - \sec \theta + 1 } = \frac { \cos \theta } { 1 - \sin \theta }

Solve: \dfrac{1+tan^2\theta}{1-tan^2\theta}=sec2\theta.\dfrac{1+tan^2\theta}{1-tan^2\theta}=sec2\theta.

Prove : \dfrac{\cot A +\text{ cosec} A -1}{\cot A - \text{cosec} A +1} = \dfrac{1 +\cos A}{\sin A}\dfrac{\cot A +\text{ cosec} A -1}{\cot A - \text{cosec} A +1} = \dfrac{1 +\cos A}{\sin A}

Prove \dfrac{1}{\sec\theta -1}-\dfrac{1}{\sec\theta +1}=2\cot^2\theta\dfrac{1}{\sec\theta -1}-\dfrac{1}{\sec\theta +1}=2\cot^2\theta

Solve \dfrac{\cos{\theta}}{1-\sin{\theta}}+\dfrac{\cos{\theta}}{1+\sin{\theta}}=4\dfrac{\cos{\theta}}{1-\sin{\theta}}+\dfrac{\cos{\theta}}{1+\sin{\theta}}=4 for \theta <90\theta <90

If \cos \theta=\dfrac{3}{5}\cos \theta=\dfrac{3}{5}, find the value of \cot \theta +cosec \ \theta\cot \theta +cosec \ \theta.

If \sin \theta =\dfrac{a}{\sqrt{a^2+b^2}}\sin \theta =\dfrac{a}{\sqrt{a^2+b^2}}, 0<\theta <90^o0<\theta <90^o, find the values of \cos \theta\cos \theta and \tan \theta\tan \theta.

The value of \cot^{2}C-\dfrac{1}{\sin^{2}C}\cot^{2}C-\dfrac{1}{\sin^{2}C} is

The value of \tan^{2}A-\sec^{2}A\tan^{2}A-\sec^{2}A is

Write \tan\theta\tan\theta in terms of \sin\theta\sin\theta.

If \sin 3A=\cos (A-10^o)\sin 3A=\cos (A-10^o), then find the value of A, where 3A is an acute angle.

Prove that \dfrac {\cot A-\cos A}{\cot A+\cos A}=\dfrac {\text{cosec} A-1}{\text{cosec} A+1}\dfrac {\cot A-\cos A}{\cot A+\cos A}=\dfrac {\text{cosec} A-1}{\text{cosec} A+1}

Eliminate \theta\theta, if x=a\cos\theta, y=a\sin \thetax=a\cos\theta, y=a\sin \theta.

Prove that :(\csc A-\sin A)(\sec A-\cos A)=\dfrac {1}{\tan A+\cot A}(\csc A-\sin A)(\sec A-\cos A)=\dfrac {1}{\tan A+\cot A}

If 7\csc\alpha-3\cot\alpha=77\csc\alpha-3\cot\alpha=7 prove that 7\cot\alpha-3\csc\alpha=37\cot\alpha-3\csc\alpha=3

If $\displaystyle \alpha , \beta , \gamma \epsilon \left ( 0, \frac{\pi }{2} \right )$ , then $\displaystyle\frac{\sin \left ( \alpha +\beta +\gamma \right )}{\sin \alpha +\sin \beta +\sin \gamma }< a$

Find a

For any $\theta$ , state the value of:
$\sin ^{2}\theta+\cos ^{2} \theta$

Prove that $\displaystyle \frac{\cos 10^{0}+\sin 10^{0}}{\cos 10^{0}-\sin 10^{0}}=\tan a^{0}$

Find a

Prove that:
$\displaystyle cot^2\frac{\pi}{6}+cosec\frac{5\pi}{6}+3tan^2\frac{\pi}{6}=6$

What is the value of $\displaystyle \left ( \tan A+\cot A \right )\sin A\times \cos A$ ?

What is the value of $\displaystyle \sin ^{6 }\theta +\cos ^{6}\theta +3\sin ^{2}\theta \cos ^{2}\theta$ ?

What is the value of $\displaystyle \sin ^{2}25^{0}+\sin ^{2}65^{0}$ ?

Prove that:
$\displaystyle \sin^2\frac{\pi}{6}+\cos^2\frac{\pi}{3}-\tan^2\frac{\pi}{4}=-\frac{1}{2}$

If $\displaystyle \cos \theta =\frac{2t}{1+t^{2}}$ , then find the value of $\displaystyle \tan \theta$ .

Write the principal value of $\cos^{-1}[\cos (680^o)]$ .

Prove that:

$\dfrac{sin\theta - 2sin^{3}\theta}{2cos^{3}\theta-cos\theta} =tan\theta$

Find the value of $\cos^{2} 12^{\circ} + \cos^{2} 78^{\circ}$

Show that $\left( 1+\cot ^{ 2 }{ \theta } \right) \left( 1-\cos { \theta } \right) \left( 1+\cos { \theta } \right) =1\quad$ .

Prove $\text{cosec}^{6}A - \cos^{6}A = 3\cot^{2}A \text{cosec}^{2} A + 1$ .

Simplify the following expression:
$\displaystyle 3 \, \cos \, x \, - \, 4 \, \sin \, x \,\cos \, x \, - \, \sin^2 \, x \, - \, 1$

Solve $\sin\theta +\sin 5\theta =\sin 3\theta; 0\leq \theta \leq \pi$ .

Solve $\sin x+\sqrt{3}\cos x \geq 1$ .

Prove the following identities :
$\dfrac{1 \, - \, \cos \, \theta}{1 \, + \,\cos \, \theta} \, = \, (cosec \, \theta \, - \, \cot \, \theta)^2$

Solve $1-2\sin \theta -2\cos \theta +\cot \theta =0$ , $(0 < \theta < 2\pi)$ .

Solve: $2+7\tan^2x=3.25sec^2x$ .

Convert $230^\circ$ into radian.

Solve: $4\sin^4x+\cos^4x=1$ .

Convert $\dfrac{3\pi}{7}$ in degrees.

$2^{\cos 2x}=3.2^{\cos^2x}-4$

The expression $\dfrac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}}$ reduces to:

Prove that:
$\cot (\pi -\theta) = -\cot \theta$

Prove that:
$\sec(\pi - \theta) = - \sec\theta$

Prove that:
$\text{cosec} (\pi -\theta)= \text{cosec }\theta$

Find the principal solution or solution of $\sin x=\dfrac {1}{\sqrt {2}}$ .

Prove that :
$\dfrac{1}{\sec \, \theta \, - \, \tan \, \theta} \, = \, \sec \, \theta \, + \, \tan \, \theta$

show that $\sqrt { 2+\sqrt { 2+\sqrt { 2+2\cos { 8\theta } } } } =2\cos { \theta }$ , $0<\theta<\dfrac{\pi}{8}$

If $\theta$ is an acute angle and $sin \theta=\cos \theta$ find $3tan\theta$

Find principal solution for $\tan x =-1$ . $x\epsilon(\dfrac{\pi}{2} ,\pi)$

Convert ${290}^{o}$ into radian measure

Find the $particular solution$ of $\tan x + \tan 2x + \tan x.\tan 2x = 1$ .

${\cos ^4}\theta + {\cos ^2}\theta = 1$ then show that ${\sec ^4}\theta - {\sec ^2}\theta = 1$

Prove that $\sum a(\sin \,B - \sin C) = 0$

Solve equation for general solution $2(\sin \, x)^2 + (\sin \, 2x)^2 = 2$ .

Find radian measure corresponding to the degree measure $- {37^ \circ }30'$

If $\sin { A } =\cfrac { 5 }{ 13 }$ then evaluate $\cos { A }$ and $\tan { A }$ .

If $\sin 9x = \sin x$ . Find the values of $x$ .

Solve $\sqrt { \cfrac { 1+\cos { 30 } }{ 1-\cos { 30 } } }$

Prove $\sin ^2\theta +\cos ^2\theta =1$

Prove if $sinA = sinB$ then $\angle A = \angle B$ .

If $\tan { 2\theta =\cot { \left( \theta +{ 6 } \right) } }$ , where $2\theta$ and $\theta +6$ are acute angles, find the value of $\theta$ .

Solve $\tan x + \cot x$

prove that:

$1+\cot^{2} \theta = \csc^{2} \theta$

Solve $\sqrt 2 \left( {\sin x + \cos x} \right) = \sqrt 3$

Find ' $x$ ' if $3\tan^{-1} x + \cot^{-1}x = \pi$ .

Solve ${\tan ^2}\theta + {\cot ^2}\theta = 2$

Solve $\tan 5\theta = \cot 2\theta$

If $\tan \theta=\dfrac{3}{A}$ , find $\cos \theta \csc \theta$ .

Solve the equations
$3(\sec^{2} \theta + \tan^{2} \theta) = 5$ .

If $co\sec{A}=2$ , find the value of $\cfrac{1}{\tan{A}}+\cfrac{\sin{A}}{1+\cos{A}}$

If $\sin{\theta}=\cfrac{1}{2}$ , show that $(3\cos{\theta}-4{\cos}^{3}{\theta})=0$

Write the complementary angle of $\dfrac{1}{3}$ of $180^o$ .

Prove that $\cos 4x=1-8\sin^2 x \cos^2x$

cos $x=\frac{1}{2}$
Find the principle solutions.

If $\sin A = \frac{1}{3}$ , then find the value of ${\mathop{\rm cosAcosec}\nolimits} + tanAsecA$

Find number of solutions of equation $\sin x=-4x+1$

Find the principal solution of the following equation:
$\cot { x=\sqrt { 3 } }$

Value of $\sin ^ { 2 } A + \cos ^ { 2 } A = ?$

If $0 < \theta < 90^o$ and $\sec \theta = \text{cosec}60^o$ , find the value of $2{\cos ^2}\theta - 1$

$tan^{-1}\dfrac{1}{3}+tan^{-1}\dfrac{1}{7}+tan^{-1}\dfrac{1}{5}+tan^{-1}\dfrac{1}{8}=\dfrac{\pi }{4}.$

Prove that: $\displaystyle \frac { \sec \theta + \tan \theta - 1 } { \tan \theta - \sec \theta + 1 } = \frac { \cos \theta } { 1 - \sin \theta }$

Solve: $\dfrac{1+tan^2\theta}{1-tan^2\theta}=sec2\theta.$

Prove : $\dfrac{\cot A +\text{ cosec} A -1}{\cot A - \text{cosec} A +1} = \dfrac{1 +\cos A}{\sin A}$

Prove $\dfrac{1}{\sec\theta -1}-\dfrac{1}{\sec\theta +1}=2\cot^2\theta$

Solve $\dfrac{\cos{\theta}}{1-\sin{\theta}}+\dfrac{\cos{\theta}}{1+\sin{\theta}}=4$ for $\theta <90$

If $\cos \theta=\dfrac{3}{5}$ , find the value of $\cot \theta +cosec \ \theta$ .

If $\sin \theta =\dfrac{a}{\sqrt{a^2+b^2}}$ , $0<\theta <90^o$ , find the values of $\cos \theta$ and $\tan \theta$ .

The value of $\cot^{2}C-\dfrac{1}{\sin^{2}C}$ is

The value of $\tan^{2}A-\sec^{2}A$ is

Write $\tan\theta$ in terms of $\sin\theta$ .

If $\sin 3A=\cos (A-10^o)$ , then find the value of A, where 3A is an acute angle.

Prove that $\dfrac {\cot A-\cos A}{\cot A+\cos A}=\dfrac {\text{cosec} A-1}{\text{cosec} A+1}$

Eliminate $\theta$ , if $x=a\cos\theta, y=a\sin \theta$ .

Prove that :
$(\csc A-\sin A)(\sec A-\cos A)=\dfrac {1}{\tan A+\cot A}$

If $7\csc\alpha-3\cot\alpha=7$ prove that $7\cot\alpha-3\csc\alpha=3$

Prove that $sec^{4}\theta-tan^{4}\theta=1+2 tan^{2}\theta.$