Class 10 Maths - Introduction To Trigonometry

Find value of 10x

Given,

$\triangle ABC$ ,

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

$\angle A = 60$ and

$BC = 20$

Now,

$\sin 60 = \frac{P}{H}$

$\sin 60 = \dfrac{BC}{AC} = \dfrac{20}{x}$

$\dfrac{\sqrt{3}}{2} = \dfrac{20}{x}$

$x = \dfrac{40}{\sqrt{3}}$

$x = 23.1$

Value of 100x is 3464,

Given,

$\triangle ABC$ ,

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

$\angle A = 30^{\circ}$ ,

$BC = 20$ and

$AB = x$

Now,

$\tan 30 = \dfrac{P}{B} = \dfrac{20}{x}$

$\dfrac{1}{\sqrt{3}} = \dfrac{20}{x}$

$x = 20 \sqrt{3}$

$x = 34.64$

$100x = 3464$

From the given figure, find the value of m if, tan C is $\displaystyle \frac{m}{4}$ .

Given:

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

$AB = 3$ and

$BC = 4$

$\triangle ABC$ ,

$\displaystyle \tan C = \frac{AB}{BC} = \frac{3}{4}$

$\displaystyle but, \ \tan C=\frac{m}{4}$

$\therefore m=3$

$\displaystyle\,\frac{1\,-\,cos\,2\,A}{sin\,2\,A}\,=\,tan\,2A$

If above statement is true then mention answer as 1, else mention 0 if false

Given,

$A = 30^{\circ}$

Now,

$\displaystyle\,\frac{1\,-\,cos\,2\,A}{sin\,2\,A}\,=\,tan\,A$

L.H.S:

$\displaystyle\,\frac{1\,-\,cos\,2\,A}{\sin\,2\,A}$

=

$\displaystyle\,\frac{1\,-\,cos\,2\,(30^0) }{\sin\,2\,(30^0)}$

=

$\displaystyle\,\frac{1\,-\,cos 60^0 }{\sin 60^0}$

=

$\displaystyle\,\frac{1 - \dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}$

=

$\displaystyle\,\dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}$

=

$\displaystyle\,\dfrac{1}{\sqrt{3}}$

R.H.S:

$\tan A = \tan 30^0 = \dfrac{1}{\sqrt{3}}$

Thus,

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

Find side $c$ of the triangle $ABC$ , given that $\displaystyle A=30^{0},B=60^{0}$ and $\displaystyle b=20\sqrt{3}$

$\displaystyle C=180^{0}-\left ( A+B \right )=90^{0}$

$\displaystyle b=AC=20\sqrt{3}$
Now

$\displaystyle \frac{AC}{AB}=\sin 60^{0}$

$\displaystyle \therefore \frac{20\sqrt{3}}{AB}=\frac{\sqrt{3}}{2}$ or

$\displaystyle \sqrt{3}AB=40\sqrt{3}$

$\displaystyle \therefore AB=40=c$

If $2\cos { \theta } =1$ and $\theta$ is acute angle, then what is $\theta$ equal to?

Given

$2\cos { \theta } =1$

$\therefore \cos { \theta } =\dfrac { 1 }{ 2 }$
We know,

$\cos { { 60 }^{ o } } =\dfrac { 1 }{ 2 }$

$\therefore \cos { \theta } =\cos { { 60 }^{ o } }$

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

Find the value of the following:
$\tan ^{ 2 }{ { 60 }^{ o } } +2\tan ^{ 2 }{ { 45 }^{ o } }$

We have,

$\tan { { 60 }^{ o } } =\sqrt { 3 }$ and

$\tan { { 45 }^{ o } } =1$

$\therefore \tan ^{ 2 }{ { 60 }^{ o } } +2\tan ^{ 2 }{ { 45 }^{ o } }$

$={ \left( \sqrt { 3 } \right) }^{ 2 }+2\cdot { \left( 1 \right) }^{ 2 }$

$=3+2=5$

Prove that $\cos { \theta } \cdot cosec \ { \theta } =\cot { \theta }$

Consider left hand side of the given equation

$\Rightarrow \cos { \theta } \cdot cosec \ { \theta } =\cos { \theta } \cdot \dfrac { 1 }{ \sin { \theta } }$

$\left( \because cosec \ { \theta } =\dfrac { 1 }{ \sin { \theta } } \right)$

$=\dfrac { \cos { \theta } }{ \sin { \theta } } =\cot { \theta } =$ RHS

$\left( \dfrac { \cos { \theta } }{ \sin { \theta } } =\cot { \theta } \right)$

$\therefore \cos { \theta } \cdot cosec \ { \theta } =\cot { \theta }$

Find the value of $\sin ^{ 2 }{ { 45 }^{ o } } +\cos ^{ 2 }{ { 45 }^{ o } }$ .

We know that

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

Here,

$\theta=45^o$

$\therefore$

$\sin ^{ 2 }{ { 45 }^{ o } } +\cos ^{ 2 }{ { 45 }^{ o } } =1$

Find the value of ${ \left( \sin { \theta } +\cos { \theta } \right) }^{ 2 }+{ \left( \sin { \theta } -\cos { \theta } \right) }^{ 2 }$

We have

${ \left( \sin { \theta } +\cos { \theta } \right) }^{ 2 }+{ \left( \sin { \theta } -\cos { \theta } \right) }^{ 2 }$

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

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

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

If $\cos{\theta}=1$ , then find the value of acute angle $\theta$ .

Given,

$\cos { \theta } =1$ ...(i)
But

$\cos { { 0 }^{ \circ } } =1$ ...(ii)

From equations (i) and (ii):

$\cos { \theta } =\cos { { 0 }^{ \circ } }$

$\therefore \theta ={ 0 }^{ \circ }$

$\sin{\left({60}^{\circ}+{30}^{\circ}\right)}=$ _______.

$\sin(60^\circ+30^\circ)=\sin(90^\circ)=1$

$(1 + \tan A \cdot \tan B)^{2} + (\tan A - \tan B)^{2} = \sec^{2}A \cdot \sec^{2}B$ .

Given:

$( 1+ \tan A \tan B)^2 + (\tan A - \tan B)^2$

$= 1 +2 \tan A \tan B + \tan ^2 A \tan ^2 B + \tan ^2 A - 2\tan A \tan B + \tan ^2 B$

$= 1+ \tan ^2 A + \tan ^2 A \tan ^2 B + \tan ^2 B$

$= \sec^2 A + \tan ^2 B(\tan ^2 A +1)$

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

$= \sec ^2 A(1+ \tan ^2 B)$

$= \sec ^2 A \sec ^2 B$

Find the value of $\sin^{2}30^{\circ} + \cos^{2} 60^{\circ}$

Solution:

$\sin ^2{30}^{\circ}+\cos ^2{60}^{\circ}=\left(\dfrac12\right)^2+\left(\dfrac12\right)^2$ where

$\left(\sin {30}^{\circ}=\dfrac12,\cos {60}^{\circ}=\dfrac12\right)$

$=\cfrac14 +\cfrac14$

$=\cfrac24$

$=\cfrac12$

Prove that $\sin^2A + \dfrac{1}{1 + \tan^2A} = 1$ .

Identites :

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

LHS =

$\sin ^{2}A+\dfrac {1}{1+\tan^{2}A}$

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

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

$=1$

$=RHS$

Hence proved

Prove that $(1 + \cot^2A) \sin^2A = 1$ .

Identity :

$cosec ^{2}A-\cot ^{2}A =1$

LHS

$= \left (1+\cot ^{2}A\right) \sin ^{2}A$

$=cosec ^{2}A \times \sin^{2}A$

$=\dfrac1{sin^2A} \times sin^2A$

$=1$

$=RHS$

Prove the following :
(iv) $\dfrac{\cos(90^{\circ} \, - \, A) \sin (90^{\circ} \, - \, A)}{\tan(90^{\circ} \, - \, A)} = \, \sin^2 \, A$

L.H.S

$=\dfrac{\cos(90^{\circ} \, - \, A) \sin (90^{\circ} \, - \, A)}{\tan(90^{\circ} \, - \, A)}$

We have ,

$\tan{A}=\dfrac{1}{\cot{A}}$

$=\cos(90^{\circ} \, - \, A) \sin (90^{\circ} \, - \, A) \, \cot(90^{\circ} \, - \, A)$

We know that,

$\cos(90-\theta)=\sin\theta$

$\sin(90-\theta)=\cos\theta$

$\cot(90-\theta)=\tan\theta$

$\therefore \cos(90^{\circ} \, - \, A) \sin (90^{\circ} \, - \, A) \, \cot(90^{\circ} \, - \, A)$

$=\sin(A) \cos (A) \, tan(A)$

$=\dfrac{\sin(A) \cos (A) \, sin(A)}{\cos(A)}$ --------------(

$\tan{A}=\dfrac{\sin(A)}{\cos{A}}$ )

$=\sin(A) \sin (A)$

$=\sin^2(A)$ =R..H.S

$\because \dfrac{\cos(90^{\circ} \, - \, A) \sin (90^{\circ} \, - \, A)}{\tan(90^{\circ} \, - \, A)}=\sin^2(A)$

Hence proved

Prove that $cosec \theta \sqrt{1 - \cos^2 \theta} = 1$ .

Identity :

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

LHS =

$cosec \theta \sqrt {\left (1-cos^{2}\theta \right)}$

$=cosec \theta \sqrt {\sin^{2}\theta}$

$=cosec \theta \sin \theta$

$=\dfrac1{sin\theta} \times sin\theta$

$=1$

$=RHS$

Evaluate the following :
$\dfrac{cos 37^{\circ}}{sin 53^{\circ}}$

We have,

$\dfrac{cos \, 37^{\circ}}{sin \, 53^{\circ}} \, = \, \dfrac{cos(90^{\circ} \, - \, 53^{\circ})}{sin \, 53^{\circ}} \, = \, \dfrac{sin \, 53^{\circ}}{sin \, 53^{\circ}} \, = \, 1$

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

Evaluate the following :
$\dfrac{\tan 10}{\cot 80}$

We have

$\tan(90-\theta)=\cot\theta$

therefore,

$\tan10=\tan(90-80)=\cot80$

To evaluate,

$\dfrac{\tan10}{\cot80}$

$=\dfrac{\cot80}{\cot80}$

$=1$

Evaluate the following :

$\dfrac{\sec 11}{\text{cosec} 79}$

We have

$\sec(90-\theta)=\csc\theta$

therefore,

$\sec11=\sec(90-79)=\csc79$

To evaluate,

$\dfrac{\sec11}{\csc79}$

$=\dfrac{\csc79}{\csc79}$

$=1$

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

LHS

$=(\sec \theta + \cos \theta) (\sec \theta -\cos \theta)$

$=\sec ^{2}\theta - \cos ^{2}\theta$

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

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

$=RHS$

Hence proved

Prove that $(\sec^2\theta - 1) (co\sec^2\theta - 1) = 1$

Identities :

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

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

LHS

$=(\sec^{2}\theta-1)(cosec ^{2}\theta - 1)$

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

$=1$ .....

$\left (\because \tan\theta=\dfrac1{\cot\theta}\right )$

$=RHS$

Hence proved

Evaluate the following :
$\dfrac{tan 54^{\circ}}{cot 36^{\circ}}$

We have,

$\dfrac{tan \, 54^{\circ}}{cot \, 36^{\circ}} \, = \, \dfrac{tan(90^{\circ} \, - \, 36^{\circ})}{cot \, 36^{\circ}} \, = \, \dfrac{cot \, 36^{\circ}}{cot \, 36^{\circ}} \, = \, 1$

$[\because \, tan(90^{\circ} \, - \, \theta) \, = \, cot \, \theta]$

Evaluate the following:

$\dfrac{2 \tan 53^{\circ}}{\cot 37^{\circ}} \, - \, \dfrac{\cot 80^{\circ}}{\tan 10^{\circ}}$

Given:

$\dfrac{2 \, \tan \, 53^{\circ}}{\cot \, 37^{\circ}} \, - \, \dfrac{\cot \, 80^{\circ}}{\tan \, 10^{\circ}} \,$

$= \, \dfrac{2 \, \tan \, 53^{\circ}}{\tan(90^{\circ} \, - \, 37^{\circ})} \, - \, \dfrac{\cot(90^{\circ} \, - \, 10^{\circ})}{\tan \, 10^{\circ}} \,$ {

$\because \tan(90-A)=\cot A,\ \ \cot(90-A)=\tan A$ }

$= \, \dfrac{2 \, \tan \, 53^{\circ}}{\tan \, 53^{\circ}} \, - \, \dfrac{\tan \, 10^{\circ}}{\tan \, 10^{\circ}} \,$

$= \, 2 \, - \, 1 \, = \, 1$

Verify that $\cot ^{ 2 }{ {60}^{o} } +\sin ^{ 2 }{ {45}^{o} } +\sin ^{ 2 }{ {30}^{o} } +\cos ^{ 2 }{ {90}^{o} } =\dfrac { 13 }{ 12 }$

${\cot ^2} {60^0}+ {\sin ^2}{45^ \circ } + {\sin ^2}{30^ \circ } + {\cos ^2}{90^ \circ }$

$= {\left( {\dfrac {1}{{\sqrt 3 }}} \right)^2} + {\left( {\dfrac {1}{{\sqrt 2 }}} \right)^2} + {\left( {\dfrac {1}{2}} \right)^2} + 0$

$= \dfrac {1}{3} + \dfrac {1}{2} + \dfrac {1}{4}$

$= \dfrac {{4 + 6 + 3}}{{12}}$

$= \dfrac {{13}}{{12}}$

If $\tan { \theta } =\dfrac { 20 }{ 21 }$ , then prove that $\dfrac { 1-\sin { \theta } +\cos { \theta } }{ 1+\sin { \theta } +\cos { \theta } } =\dfrac { 3 }{ 7 }$

$\tan \theta=\dfrac{20}{21}$

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

$1+\dfrac{20\times 20}{21\times 21}=\sec^{2}\theta$

$1+\dfrac{400}{441}=\sec^{2}\theta$

$\dfrac{841}{441}=\sec^{2}\theta$

$\sec\theta =\dfrac{29}{21}$

$\cos\theta=\dfrac{21}{29}............... (1)$

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

$1=\dfrac{21\times 21}{29\times 29}+\sin^{2}\theta$

$1=\dfrac{441}{841}+\sin^{2}\theta$

$1-\dfrac{441}{841}=\sin^{2}\theta$

$\dfrac{400}{841}=\sin^{2}\theta$

$\dfrac{20}{29}=\sin\theta................(2)$

$\dfrac{1-\sin \theta +\cos\theta}{1+\sin\theta +\cos\theta}............(3)$

Substituting equation

$(1)$ and

$(2)$ in

$(3)$ ,

$\dfrac{1-\dfrac{20}{29} +\dfrac{21}{29}}{1+\dfrac{20}{29} +\dfrac{21}{29}}=\dfrac{1+\dfrac{1}{29}}{1+\dfrac{41}{29}}=\dfrac{\dfrac{30}{29}}{\dfrac{70}{29}}=\dfrac{30}{70}=\dfrac{3}{7}$

$LHS=RHS$

Hence proved.

Express each of the following in trigonometric ratios of angles between $0^{\circ} \, and \, 45^{\circ}$ :
$sin 85^{\circ} \, + \, cosec 85^{\circ}$

We have,

$sin \, 85^{\circ} \, + \, cosec \, 85^{\circ}$

$= \, sin(90^{\circ} \, - \, 5^{\circ}) \, + \, cosec(90^{\circ} \, - \, 5^{\circ})$

$= \, cos \, 5^{\circ} \, + \, sec \, 5^{\circ}$

$[\because \, sin(90^{\circ} \, - \, \theta) \, = \, cos \, \theta \, and \, cosec(90^{\circ} \, - \, \theta) \, = \, sec \, \theta]$

${ \left( \tan { x } +\cot { x } \right) }^{ 2 }=\sec ^{ 2 }{ x+\csc ^{ 2 }{ x } }$

From the properties of triangles,

$\Rightarrow 1+\tan^{2}x=\sec^{2}x$

$\Rightarrow 1+\cot^{2}x=\text{cosec}^{2}x$

Given,

$LHS={ ( \tan { x } +\cot { x } ) }^{ 2 }$

${ ( \tan { x } +\cot { x } ) }^{ 2 }=\tan^{2} x+ \cot^{2} x+2\tan x\cot x=\tan^{2} x+ cot^{2} x+2$

$=1+\tan^{2}x +1+\cot^{2} x=\sec^{2}x +\ cosec ^{2} x=RHS$

$\therefore LHS=RHS$

Hence proved.

Prove that: $\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos\,A}$

Simplifying RHS

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

Calculate : ${ \left( \cos { 40 } \right) }^{ 2 }+{ \left( \sin { 40 } \right) }^{ 2 }$

$={ (\cos 40) }^{ 2 }+{ (\sin 40) }^{ 2 }$

$=\cos ^{ 2 }{ 40 } +\sin ^{ 2 }{ 40 }$

[Since,

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

$=1$

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

Consider ${\dfrac{\cot A - \cos A} {\cot A + \cos A}},$

$\dfrac{\dfrac{\cos A}{\sin A}-\cos A}{\dfrac{\cos A}{\sin A}+\cos A}=\dfrac{\cos A-\cos A\sin A}{\cos A+\cos A\sin A }$

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

$=\dfrac{\dfrac{1}{\sin A}-\dfrac{\sin A}{\sin A}}{\dfrac{1}{\sin A}+\dfrac{\sin A}{\sin A}}$

$= {\dfrac{\cos ecA - 1}{\cos ecA + 1}}$

Prove that: $\dfrac{1}{\sec A-1}+\dfrac{1}{\sec A+1}=2 \csc A \cot A$

L.H.S

$=\dfrac{1}{\sec A-1}+\dfrac{1}{\sec A+1}$

$=\dfrac{\sec A+1+\sec A-1}{\left( \sec A-1 \right)\left( \sec A+1 \right)}$

$=\dfrac{2\sec A}{\left( {{\sec }^{2}}A-1 \right)}$

$=\dfrac{2\sec A}{{{\tan }^{2}}A}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \because {{\sec }^{2}}A-1={{\tan }^{2}}A \right]$

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

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

$=2\csc A\cot A\,\,\,\,\,$

Hence, this is the answer.

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

L.H.S

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

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

$\Rightarrow \dfrac{1+\cos \theta }{\left( 1-\cos \theta \right)\left( 1+\cos \theta \right)}$

$\Rightarrow \dfrac{1}{\left( 1-\cos \theta \right)}\,\,$

R.H.S

Hence, proved.

Evaluate
$\dfrac{tan 45^o}{cosec 30^o} + \dfrac{sec 60^o}{cot 45^o} - \dfrac{5 sin 90^o}{2 cos 0^o}$

$\dfrac{tan 45^o}{cosec 30^o} + \dfrac{sec 60^o}{cot 45^o} - \dfrac{5 sin 90^o}{2 cos 0^o}$

As we know all the values which is

given here.

$= \dfrac{1}{2}+\dfrac{2}{1}-\dfrac{5}{2}$

$= \dfrac{5}{2}-\dfrac{5}{2}$

$= 0$

Evaluate : $\sin^{2}90.\cos^{2} 45+4 \tan^{2}30+\dfrac {1}{2}\sin^{2}a-2 \cos^{2}90+\dfrac {1}{24}$

Consider the given expression.

$\Rightarrow {{\sin }^{2}}{{90}^{0}}\cos^2 {{45}^{0}}+4{{\tan }^{2}}30+\dfrac{1}{2}{{\sin }^{2}}a-2{{\cos }^{2}}{{90}^{0}}+\dfrac{1}{24}$

Put the value of all trigonometry function in given expression, we get

$\Rightarrow {{\sin }^{2}}{{90}^{0}}\cos^2 {{45}^{0}}+4{{\tan }^{2}}30+\dfrac{1}{2}{{\sin }^{2}}a-2{{\cos }^{2}}{{90}^{0}}+\dfrac{1}{24}$

$\Rightarrow {{\left( 1 \right)}^{2}}.{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}+4.{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}+\dfrac{1}{2}{{\sin }^{2}}a-2{{\left( 0 \right)}^{2}}+\dfrac{1}{24}$

$\Rightarrow \,\dfrac{1}{2}+\dfrac{4}{3}+\dfrac{1}{2}{{\sin }^{2}}a+\dfrac{1}{24}$

$\Rightarrow \,\dfrac{1}{2}{{\sin }^{2}}a+\dfrac{12+32+12}{24}$

$\Rightarrow \,\,\dfrac{1}{2}{{\sin }^{2}}a+\dfrac{56}{24}$

$\Rightarrow \dfrac{1}{2}{{\sin }^{2}}a+\dfrac{7}{3}$

Hence, this is the answer.

Simplify:
$\dfrac {\sin \theta.\csc (90-\theta).\tan \theta}{\csc (90-\theta).\cos \theta.\cot (90-\theta)}$

1325341_1086847_ans_b7100a7e19d548a28cf836c524a6fc6f.jpg

For any angle of measure $\theta$
prove that $1+\tan^2\theta =\sec^2\theta$ .

$\begin{array}{l} \sin \theta = \dfrac{{perpendicular}}{{hypo}} = \frac{P}{H}\\ \cos \theta = \dfrac{{base}}{{hypo}} = \frac{B}{H}\\ \therefore paythaguras\\ {H^2} = {P^2} + {B^2}\\ \dfrac{{{H^2} = {P^2} + {B^2}}}{{{B^2}}}\\ \dfrac{{{H^2}}}{{{B^2}}} = \dfrac{{{p^2}}}{{{B^2}}} + \dfrac{{{B^2}}}{{{B^2}}}\\ 1 + {\tan ^2}\theta = {\sec ^2}\theta \end{array}$

Find the value of $\dfrac{\cos 30^o \tan 60^o + \sin 60^o \cos 60^o}{cosec \,30^o \sec 60^o - cosec \,60^o \cot 30^o}$

$\dfrac{\cos 30^0 \tan 60^0+\sin 60^0\cos 60^0}{cosec \ 30^0\sec 60^0-coses \ 60^0\cot 30^0}$

Putting standard trigonometric values, we get

$=\dfrac{\sqrt{3}/2}{2}=\dfrac{\sqrt{3}+\sqrt{3}/2\, \, 1/2}{2-2/\sqrt{3}\sqrt{3}}$

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

$=\dfrac{6+\sqrt{3}}{4.2}=\dfrac{6+\sqrt{3}}{8}$

$\therefore \boxed{\dfrac{\cos 30^0 \tan 60^0+\sin 60^0\cos 60^0}{cosec\ 30^0\sec 60^0-cosec \ 60^0\cot 30^0}=\dfrac{6+\sqrt{3}}{8}}$

Prove that: $\sqrt {\sec^{2}A+\csc^{2}A}=\tan A+\cot A$

$LHS$

$=\sqrt{{{\sec }^{2}}A+{{\csc }^{2}}A}$

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

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

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

$=\csc A\sec A.$

Now,

$RHS$

$=\tan A+\cot A$

$=\dfrac{\sin A}{coA}+\dfrac{\cos A}{\sin A}$

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

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

$=\csc A\sec A$

$=LHS.$

If $A = 15^o$ , verify that $4 \sin 2 A . \cos 4 A . \sin 6 A = 1$

$A=15^{\circ}$

Therefore,

$4\sin 2A \cos 4A \sin 6A$

$=4\sin 30^{\circ} \cos 60^{\circ} \sin 90^{\circ}$

$=4\times \dfrac{1}{2}\times \dfrac{1}{2}\times 1$

$=1$

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

$sin^2\theta(tan^2\theta+1)=sec^2\theta-1$

$\Rightarrow$

$sin^2\theta.sec^2\theta=sec^2\theta-1$

$\Rightarrow$

$sin^2\theta.sec^2\theta-sec^2\theta=-1$

$\Rightarrow$

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

LHS,

$\Rightarrow$

$sec^2\theta(-cos^2\theta)$

$\Rightarrow$

$-1$ = RHS

Hence proved

Evaluate
$\tan\theta.\sin\theta+\cos\theta=\sec\theta$

Changing LHS into sine, cosine,

$\Rightarrow$

$\dfrac{sin^2\theta}{cos\theta}+cos\theta$

Taking LCM,

$\Rightarrow$

$\dfrac{sin^2\theta+cos^2\theta}{cos\theta}$

$\Rightarrow$

$\dfrac{1}{cos\theta}$

$\Rightarrow$

$sec\theta$ = RHS

Hence proved.

Simplify: $\dfrac{{{\text{sec}}{{\text { x}}^{\text{}}}{\text{ + tan}}{{\text{ x}}^{\text{}}}}}{{{\text{sec}}{{\text{ x}}^{\text{}}}{\text{ - tan}}{{\text { x}}^{\text{}}}}}$

$\dfrac { \sec x + \tan x } { \sec x - \tan x }$

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

$= \dfrac { 1 + \sin x } { 1 - \sin x }$

In the figure, find the value of $BC$

$BE = 100\, \cos\,60^{\circ}= 100\times \dfrac{1}{2} = 50\ \text{m}$

$\tan \angle DEC = \tan 45^{\circ} = \dfrac{DC}{EC} = 1$

$DC = 80\ \text{m} = EC$

$\Rightarrow BC = BE+EC = 50\ \text{m}+80\ \text{m} = 130\ \text{m}$

$\therefore BC = 130\ \text{m}$

Solve $\dfrac{{\cos {{45}^ \circ }}}{{\sec {{30}^ \circ } + {\mathop{\rm cosec}\nolimits} {{30}^ \circ }}}$

$\dfrac { \cos { { 45 }^{ ° } } }{ \sec { { 30 }^{ ° } } +cosec{ { 30 }^{ ° } } } \\ =\dfrac { \dfrac { 1 }{ \sqrt { 2 } } }{ \dfrac { 2 }{ \sqrt { 3 } } +\dfrac { 2 }{ 1 } } \\ =\dfrac { \sqrt { 3 } }{ 2\sqrt { 2 } (\sqrt { 3 } +1) }$

Find value of:
$3\cos ^{ 2 }{ { 30 }^{ o } } +\sec ^{ 2 }{ { 30 }^{ o } } +2\cos ^{ 2 }{ { 0 }^{ o } } +3\sin ^{ 2 }{ { 90 }^{ o } } -\tan ^{ 2 }{ { 60 }^{ o } }$

We know the value of

$cos30^o=\dfrac{\sqrt{3}}{2}, sec30^o=\dfrac{2}{\sqrt{3}}, cos0^o=1,sin90^o=1,tan60^o=\sqrt{3}$ . Substituting these values in the expression given in the question, we obtain,

$3.\frac{3}{4}+\frac{4}{3}+2+3-3=\frac{67}{12}$

Evaluate: $\dfrac{\cos\ 3A-2\cos(4A)}{\sin\ 3A+2\sin(4A)}$ ; when $A=15^{o}$

1263875_1318697_ans_27149c2ac5ed44c2842006a02017a1a0.jpeg

Solve:-
$\dfrac{{2\tan {{45}^ \circ }}}{{1 + {{\tan }^2}{{45}^ \circ }}}$

$tan 45^{\circ}=1$

$\Rightarrow \dfrac{2\times 1}{1+1^{2}}=\dfrac{2}{2}=1$

Prove that $\cos \theta \sin(90^o-\theta)+\sin \theta \cos(90^o-\theta)=1$

L.H.S.

$\cos \theta \sin (90^o-\theta) + \sin \theta \cos (90^o-\theta)$

$= \cos \theta. \cos \theta + \sin \theta. \sin \theta$

$= \cos^2 \theta + \sin^2\theta = 1 =$ R.H.S.

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

$\cos \theta \sin (90^o-\theta) + \sin \theta \cos (90^o-\theta)=1$

Hence Proved

Without using trigonometric tables, evaluate the following: $\left ( \sin 35^{0}\cos 55^{0}+\cos 35^{0}\sin 55^{0} \right )/\left ( cosec^{2}10^{0}-\tan ^{2}80^{0} \right )$ .

$\left ( \sin 35^{0}\cos 55^{0}+\cos 35^{0}\sin 55^{0} \right )/\left ( cosec^{2}10^{0}-\tan ^{2}80^{0} \right )$

$\frac{\sin 35^{0}\cos \left ( 90^{0}-35^{0} \right )+\cos 35^{0}\sin \left ( 90^{0}-35^{0} \right )}{cosec^{2}10^{0}-\tan ^{2}\left ( 90^{0}-10^{0} \right )}$

$\frac{\sin 35^{0}\sin 35^{0}+\cos 35^{0}\cos 35^{0}}{cosec^{2}10^{0}-\cot ^{2}10^{0}}$

$\frac{\sin ^{2}35^{0}+\cos ^{2}35^{0}}{cosec^{2}10^{0}-\cot ^{2}10^{0}}$

$\frac{1}{1}=1$ .

Prove that following identities, where the angles involved are acute angles for which the trigonometric ratios as defined: $\left ( 1+\tan ^{2}A \right )\left ( 1-\sin A \right )\left ( 1+\sin A \right )=1$ .

$L.H.S.=\left ( 1+\tan ^{2}A \right )\left ( 1-\sin A \right )\left ( 1+\sin A \right )$

$\left ( 1+\frac{\sin ^{2}A}{\cos ^{2}A} \right )\left ( 1-\sin ^{2}A \right )$

$\frac{\cos ^{2}A+\sin ^{2}A}{\cos ^{2}A}*\cos ^{2}A$

$\frac{1}{\cos ^{2}A}=\cos ^{2}A=1=R.H.S$

$\left \{ \because 1-\sin ^{2}A=\cos ^{2}A \right \}$

$\left \{ \sin ^{2}+\cos ^{2}A=1 \right \}$ .

Verify that
$\tan^{2} 30^{\circ} + \tan^{2} 45^{\circ} + \tan^{2} 60^{\circ} = 4\dfrac {1}{3}$ .

$\tan^{2} 30^{\circ} =\frac{1}{3}$

$\tan^{2} 45^{\circ} =1$

$\tan^{2} 60^{\circ} = 3$

$\tan^{2} 30^{\circ} + \tan^{2} 45^{\circ} + \tan^{2} 60^{\circ} = \frac{1}{3}+1+3$ .

$4\dfrac {1}{3}$ .

A $15 m$ long ladder touches the top of vertical wall. If this ladder makes an angle of $60^{0}$ with the wall, then height of the wall is :

REF.Image

$sin60^{\circ}=\frac{AB}{AC}=\frac{AB}{15}$

$AB=15 sin 60^{\circ}=\frac{15\sqrt{3}}{2}m$

1190733_1426668_ans_1685ea7adade48cf8592bb3fecbcf6d9.jpg

Prove that following identities, where the angles involved are acute angles for which the trigonometric ratios as defined: $1/\left ( 1+\cos A \right )+1/\left ( 1-\cos A \right )=2cosec^{2}A$ .

$L.H.S.=1/\left ( 1+\cos A \right )+1/\left ( 1-\cos A \right )$

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

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

$\left \{ \because 1-\cos ^{2}A=\sin ^{2}A \right \}$

$2cosec^{2}A=R.H.S.$ .

Verify that
$\cos 45^{\circ} \cos 60^{\circ} - \sin 45^{\circ} \sin 60^{\circ} = -\dfrac {\sqrt {3} - 1}{2\sqrt {2}}$ .

1662497_1731224_ans_b928049f20374ab09a6bcd691c189cb0.PNG

Verify that
$cosec^{2} 45^{\circ} \cdot \sec^{2} 30^{\circ} \cdot \sin^{3} 90^{\circ} \cdot \cos 60^{\circ} = 1\dfrac {1}{3}$ .

$cosec^{2} 45^{\circ}=2$

$\sec^{2} 30^{\circ}=\frac{4}{3}$

$\sin^{3} 90^{\circ} =1$

$\cos 60^{\circ}=\frac{1}{2}$

$cosec^{2} 45^{\circ} \cdot \sec^{2} 30^{\circ} \cdot \sin^{3} 90^{\circ} \cdot \cos 60^{\circ}=2*\frac{4}{3}*1*\frac{1}{2}=\frac{4}{3}$

Find the value of $\cfrac{\tan 47^{0}}{\cot 43^{0}}$

$\textbf{Evaluate the given expression using trigonometric properties}$

$\text{Given expression is,}$

$\Rightarrow\dfrac{\tan 47^{\circ}}{\cot 43^{\circ}}$

$\Rightarrow\dfrac{\tan\left(90^{\circ}- 43^{\circ}\right)}{\cot 43^{\circ}}$

$\boldsymbol{\left[\because\tan\left(90^{\circ}-\theta\right)=\cot\theta\right]}$

$\Rightarrow\dfrac{\cot 43^{\circ}}{\cot 43^{\circ}}$

$=1$

$\textbf{Hence the value of given expression is 1}$

Simplify: $(1 + \tan^{2} \theta) (1 - \sin \theta) (1 + \sin \theta)$ .

$(1 + \tan^{2} \theta) (1 - \sin \theta) (1 + \sin \theta)$

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

$=(\sec^2\theta)(\cos^2\theta)$

$=(\sec^2\theta)(\dfrac{1}{\sec ^2\theta})$

$=1$

State true or false:
$\sin^2A+ \cos^2A= 1$

$\sin^2A+\cos^2A=1$ . Its a standard identity.

We know that

$\sin A = \dfrac{P}{H}$ and

$\cos A = \dfrac{B}{H}$

Hence

$\sin ^{2} A = \dfrac{P^{2}}{H^{2}}$ and

$cos^{2}A = \dfrac{B^{2}}{H^{2}}$

Hence adding the above given functions

$sin^{2}A + cos ^{2}A$ we get

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

Using

$P^{2}+B^{2} = H^{2}$ by Pythagoras Theorem

If $\sin A, \cos A$ and $\tan A$ are in geometric progression, then $\cot^6A-\cot^2A=........$

sice $\sin A. \cos A$ and $\tan A$ are G.P. we have $\cos ^2A=\sin A \tan A$
$\Rightarrow \cot ^2A=sec A \Rightarrow \cot ^4A=1+\tan ^2A$
$\Rightarrow \cot ^6A-\cot ^2A=1$

$\tan 30^{\circ} \: \cot 30^{\circ}=$

$\tan30^{\circ} \cot 30^{\circ}$

$=\dfrac{1}{\sqrt{3}} \times \sqrt{3}$

$=1$

$\text{cosec}\, ^{2} 60^{0}-\tan ^{2} 30^{0}=$

$\csc ^{2} 60^{0}-\tan ^{2} 30^{0}$

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

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

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

$=1$

If the angle $\theta = 60^{\circ}$ and find the value of sec $\theta$ .

$\theta = 60^{\circ}$

Hence,

$\sin \theta = \sin60^{\circ} = \dfrac{\sqrt{3}}{2}$

$\cos \theta = \cos 60^{\circ} = \dfrac{1}{2}$

$\tan\theta = \tan60^{\circ} = \sqrt{3}$

$\sec \theta = \sec 60^{\circ} = 2$

$\displaystyle \dfrac{2 \tan 53^o}{\cot 37^o} - \dfrac{\cot 80^o}{\tan 10^o}$

$\displaystyle \dfrac{2 \tan (90 - 37)^o}{\cot 37^o} - \dfrac{\cot (90 - 10)^o}{\tan 10^o}$

$\displaystyle \dfrac{2 \cot 37^o}{\cot 37^o} - \dfrac{\tan 10^o}{\tan 10^o}$

$2 - 1$
=

$1$

If $\displaystyle \sin \theta = \cfrac{\sqrt{2}}{3}, \cos \theta = \cfrac{1}{3}$ , then $\displaystyle \tan \theta = \sqrt{m}.$ The value of $m$ is

$\tan \theta =\dfrac{\sin \theta}{\cos \theta}=\dfrac{\dfrac{\sqrt2}{3}}{\dfrac{1}{3}}=\sqrt2$

\sin ce,

$\tan \theta=\sqrt m=\sqrt2\Rightarrow m=2$

Therefore, Answer is

$2$

If $\displaystyle tan \theta = \frac{2}{5}$ , then $\displaystyle cot \theta = \frac{p}{2}$ , $p$ is

Given:

$tan \theta = \cfrac{2}{5}$
Then,

$\cot \theta = \cfrac{1}{tan\theta} = \cfrac{1}{\cfrac{2}{5}} = \cfrac{5}{2}$

$\therefore p=5$

If $\tan 2A = \cot (A - 18^o)$ then find the value of A where (2A) and $(A - 18^o)$ are acute angles

$\tan2A = \cot(A - 18)$

$=>\cot(90 - 2A) = \cot(A - 18)$

$=>90 - 2A = A - 18$

$=>3A = 108$

$=>A = 36^{\circ}$

The value of $\displaystyle \left ( \dfrac{\tan 60^{0}+1}{\tan 60^{0}-1} \right )^{2}= \dfrac{1+\cos 30^{0}}{1-\cos 30^{0}}$

if true then enter $1$ and if false then write $0$

$\displaystyle \left ( \frac{\tan 60^{0}+1}{\tan 60^{0}-1} \right )^{2}= \frac{1+\cos 30^{0}}{1-\cos 30^{0}}$

L.H.S:

$\displaystyle \left ( \frac{\tan 60^{0}+1}{\tan 60^{0}-1} \right )^{2}$

$\displaystyle \left ( \frac{\sqrt{3}+1}{\sqrt{3} - 1} \right )^{2}$

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

$\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}}$

R.H.S:

$\dfrac{1+\cos 30^{0}}{1-\cos 30^{0}}$

$\displaystyle \dfrac{1+\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{2}}$

$\displaystyle \dfrac{2 + \sqrt{3}}{2 - \sqrt{3}}$

Thus,

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

The value of $\displaystyle \tan \left ( 2 \times 30^{0}\right )= \frac{2 \tan 30^{0}}{1-\tan ^{2} 30^{0}}$

if true then enter $1$ and if false then write $0$

To Prove:

$\displaystyle \tan \left ( 2 \times 30^{0}\right )= \frac{2 \tan 30^{0}}{1-\tan ^{2} 30^{0}}$

L.H.S:

$tan ( 2 \times 30^{\circ}) = tan 60^{\circ} = \sqrt{3}$
R.H.S:

$\frac{2 \tan 30^{0}}{1-\tan ^{2} 30^{0}}$
=

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

$\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$
=

$\sqrt{3}$
Thus,

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

The value of $\cos ^{2} 30^{0}-\sin ^{2} 30^{0}= \cos 60^o$
If above statement is true then enter $1$ and if false then write $0$ .

$\cos ^{2} 30^{0}-\sin ^{2} 30^{0}= \cos 60^{0}$

L.H.S.:

$=\cos ^{2} 30^{0}-\sin ^{2} 30^{0}$

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

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

$\dfrac{2}{4}$

$\dfrac{1}{2}$

$\cos 60^{\circ}$

hence, L.H.S. = R.H.S.

$\sin ^{2} 30^{0}+\cos ^{2} 30^{0}$

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

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

$=\dfrac{4}{4}$

$=1$

The value of $\cos 30^{\circ} \cos 60^{\circ}-\sin 30^{\circ} \sin 60^{\circ}$ is equal to

$\begin{aligned}{} \cos {30^\circ}\cos {60^\circ} - \sin {30^\circ}\sin {60^\circ} &= \frac{{\sqrt 3 }}{2} \times \frac{1}{2} - \frac{1}{2} \times \frac{{\sqrt 3 }}{2}\\& = \frac{{\sqrt 3 }}{4} - \frac{{\sqrt 3 }}{4}\\ &= 0\end{aligned}$

Hence proved, given expression is equal to

$0.$

$ABC$ is an isosceles right-angled triangle. Assuming $AB= BC= x$ , find the value of each of the following trigonometric ratios:
$\sin 45^{0}$ is $\displaystyle \frac{1}{\sqrt{2}}$
If true then enter $1$ and if false then enter $0$

Given:

$\triangle ABC$ ,

$AB = BC = x$ ,

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

By Pythagoras Theorem,

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

$AC^2 = x^2 + x^2$

$AC^2 = 2x^2$

$AC = \sqrt{2} x$

Since,

$AB = BC$ , then by Isosceles triangle property,

$\angle ACB = \angle CAB = 45^{\circ}$

Now,

$Sin 45^{\circ} = Sin C = \frac{AB}{AC} = \frac{x}{x\sqrt{2}} = \frac{1}{\sqrt{2}}$

Find the value of $\text{cosec}^{2} 45^{o}-\cot^{2} 45^{o}$ .

To find:

$\text{cosec}^{2} 45^{o}-\cot^{2} 45^{o}$

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

$=2 - 1$

$=1$

Hence,

$\text{cosec}^{2} 45^{o}-\cot^{2} 45^{o}=1$

The value of $\displaystyle \cos \left ( 2 \times 30^{0}\right )= \frac{1-\tan ^{2} 30^{0}}{1+\tan ^{2} 30^{0}}$

If true then enter $1$ and if false then write $0$

To prove:

$\displaystyle \cos \left ( 2 \times 30^{0}\right )= \frac{1-\tan ^{2} 30^{0}}{1+\tan ^{2} 30^{0}}$

L.H.S :

$cos (2 \times 30^{\circ}) = cos 60^{\circ} = \frac{1}{2}$
R.H.S.:

$\frac{1-\tan ^{2} 30^{0}}{1+\tan ^{2} 30^{0}}$
=

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

$\frac{\frac{2}{3}}{\frac{4}{3}}$
=

$\frac{1}{2}$
Thus,

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

The value of $\displaystyle \sin \left ( 2 \times 30^{0}\right )= \frac{2\tan 30^{0}}{1+\tan ^{2} 30^{0}}$

if true then enter $1$ and if false then write $0$

$\displaystyle \sin \left ( 2 \times 30^{0}\right ) = \frac{2\tan 30^{0}}{1+\tan ^{2} 30^{0}}$

L.H.S :

$\displaystyle \sin \left ( 2 \times 30^{0}\right )$
=

$\displaystyle \sin \left (60^{0}\right )$
=

$\displaystyle \frac{\sqrt{3}}{2}$

R.H.S:

$\displaystyle \frac{2\tan 30^{0}}{1+\tan ^{2} 30^{0}}$
=

$\displaystyle \frac{2 \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$
=

$\displaystyle \frac{2 \frac{1}{\sqrt{3}}}{\frac{3 + 1}{3}}$
=

$\displaystyle \frac{2 \frac{1}{\sqrt{3}}}{\frac{4}{3}}$
=

$\displaystyle \frac{\sqrt{3}}{2}$

Thus,

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

If $A= 30^{0}$ , then the value of
$\sin \left ( 60^{0}+A \right )-\sin \left ( 60^{0}-A \right )= \sin A$
If true then enter $1$ and if false then write $0$

Given,

$A = 30^{\circ}$

To Prove:

$\sin \left ( 60^{0}+A \right )-\sin \left ( 60^{0}-A \right )= \sin A$

$L.H.S=\sin \left ( 60^{0}+A \right )-\sin \left ( 60^{0}-A \right )$
=

$\sin \left ( 60^{\circ}+30^{\circ} \right )-\sin \left ( 60^{\circ}-30^{\circ} \right )$
=

$\sin \left ( 90^{\circ} \right )-\sin \left ( 30^{\circ}\right )$
=

$1 - \dfrac{1}{2}$
=

$\dfrac{1}{2}$

$R.H.S=sin A = sin 30^{\circ} = \dfrac{1}{2}$
Thus,

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

If $\tan \theta= \cot \theta$ and $0^{0}\leq \theta \leq 90^{0}$ , state the value of $\theta$ in degrees

$\tan \theta = \cot \theta$

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

$\tan ^2 \theta = 1$

$\tan \theta = \pm 1$

Neglecting the negative value,

$\tan \theta = 1$

$\tan \theta = \tan 45$

$\theta = 45^{\circ}$

The value of $\sin 60^{0}=2 \sin 30^{0} \cos 30^{0}$

if true then enter $1$ and if false then write $0$

To Prove:

$\sin 60^{0}=2 \sin 30^{0} \cos 30^{0}$

L.H.S.:

$sin 60^{\circ} = \frac{\sqrt{3}}{2}$
R.H.S.:

$2 \sin30^{\circ} \cos 30^{\circ} = 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$
Thus,

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

If $\sqrt{3}= 1.732$ , (correct to two decimal places) the value of $\sin 60^{0}$ is $0.87.$
If true then enter $1$ and if false then enter $0$ .

Given,

$\sqrt 3=1.732$ .

Now,

$\sin 60^o\\$

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

$=\dfrac{1.732}{2}\\$

$=0.866\\$

$=0.87$ .

So the statement is true.

So the correct answer is

$1$ .

If $x= 15^{0}$ , evaluate:
$\displaystyle 12 \sin 3x \cos 3x+5 \tan ( 2x+ 15^{0})-3 \cot ^{2}3x$

Given,

$x = 15^{\circ}$
Then,

$\displaystyle 12 \sin 3x \cos 3x+5 \tan ( 2x+ 15^{0})-3 \cot ^{2}3x$
=

$\displaystyle 12 \sin 45^0 \cos 45^0 + 5 \tan ( 30^0+ 15^{0})-3 \cot ^{2}45^0$
=

$\displaystyle 12 (\frac{1}{\sqrt{2}}) (\frac{1}{\sqrt{2}}) + 5 (1)- 3 (1)^2$
=

$\displaystyle 6 + 5 - 3$
=

$8$

If $\sin x= \cos x$ and $x$ is acute, state the value of $x$ in degrees.

$\sin x = \cos x$

$\tan x = 1$

$\tan x = \tan 45$
Thus,

$x = 45^{\circ}$

If $A= 60^{0}$ and $B= 30^{0}$ , find the value of $\left ( \sin A \cos B+ \cos A\sin B \right )^{2}+\left ( \cos A \cos B-\sin A \sin B \right )^{2}$

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

When , $A = 60^{\circ}$ and $B = 30^{\circ}$

= $\left ( \sin 60 \cos 30 + \cos 60\sin 30 \right )^{2}+\left ( \cos 60 \cos 30-\sin 60 \sin 30 \right )^{2}$

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

= $\left ( \dfrac{3}{4} + \dfrac{1}{4} \right )^{2}$

= $1^2$

= $1$

If $x= 15^{0}$ , evaluate:
$4 \cos 2x \sin 4x\tan 3x-1$

Given,

$x = 15^{\circ}$

Now,

$4 \cos 2x \sin 4x\tan 3x-1$
=

$4 (\cos 30^0) (\sin 60^0) (\tan 45^0) -1$
=

$4 (\dfrac{\sqrt{3}}{2}) (\dfrac{\sqrt{3}}{2}) (1) - 1$
=

$3 - 1$
=

$2$

If $\sqrt{3}= 1.732$ , find (correct to two decimal places) the value of the following:

$\displaystyle \dfrac{2}{\tan 30^{0}}$ is $3.46$

If true then enter $1$ and if false then enter $0$

It is given that

$\sqrt{3}= 1.732$

Therefore,

$\dfrac{2}{\tan 30^{\circ}} = \dfrac{2}{\dfrac{1}{\sqrt{3}}}$

$= 2 \sqrt{3}$

$= 3.46$

If $A= 30^{0}$ , then
$\displaystyle \cos 2A= \cos ^{2} A-\sin ^{2} A= \dfrac{1-\tan ^{2}A}{1+\tan ^{2}A}$
If true enter 1 else 0

Given,

$A = 30^{\circ}$

Now,

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

L.H.S :

$\displaystyle \cos 2A = \cos 2 \times 30 = \cos 60 = \frac{1}{2}$

Middle Side:

$\cos^2{A} - sin^2{A} = \cos^2{30} - sin^2{30} = (\frac{\sqrt{3}}{2})^2 - (\frac{1}{2})^2$
=

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

$\frac{2}{4}$
=

$\frac{1}{2}$

R.H.S:

$\frac{1-\tan ^{2}A}{1+\tan ^{2}A}$
=

$\frac{1-\tan ^{2} 30}{1+\tan ^{2} 30}$
=

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

$\frac{\frac{2}{3}}{\frac{4}{3}}$
=

$\frac{1}{2}$

Hence,

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

$\displaystyle \tan \left ( A-B \right )= \dfrac{\tan A-\tan B}{1+\tan A \tan B}$
If true enter 1 else 0

Given,

$A= 60^{\circ}, B = 30^{\circ}$

Now,

$\displaystyle \tan \left ( A-B \right )= \frac{\tan A-\tan B}{1+\tan A \tan B}$

L.H.S:

$\tan (A - B) = \tan (60 - 30) = \tan 30 = \frac{1}{\sqrt{3}}$

R.H.S:

$\dfrac{\tan A-\tan B}{1+\tan A \tan B}$

$= \dfrac{\tan 60 -\tan 30}{1+\tan 60 \tan 30}$

$=\dfrac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + \sqrt{3} \times\frac{1}{\sqrt{3}}}$

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

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

Thus,

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

If $A= 30^{0}$ , then
$2\cos ^{2} A-1= 1-2\sin ^{2}A$
If true enter 1 else 0

$A = 30^{\circ}$

Now,

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

L.H.S:

$2 cos^{2} A -1 = 2 cos^{2} 30 -1$
=

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

$2 (\frac{3}{4}) - 1$
=

$\frac{1}{2}$

R.H.S :

$1 - 2\sin ^{2}A$
=

$1 - 2 \sin^2 {30}$
=

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

$1 - \frac{1}{2}$
=

$\frac{1}{2}$

Thus,

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

If $A=B= 45^{0}$ , then
$\sin \left ( A-B \right )=\sin A \cos B-\cos A \sin B$
If true enter 1 else 0

Given,

$A = B = 45^{\circ}$

Now,

$\sin \left ( A-B \right )=\sin A \cos B-\cos A \sin B$

L.H.S:

$\sin (A - B) = \sin (45^0 - 45^0) = \sin 0^0 = 0$

R.H.S :

$\sin A \cos B-\cos A \sin B$
=

$\sin 45^0 \cos 45^0 -\cos 45^0 \sin 45^0$
=

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

$\frac{1}{2} - \frac{1}{2}$
=

$0$

$= \csc (A-20^{\circ}$ )

$\sec 4A =\csc(A - 20)$
or

$\cos 4A = \sin(A- 20)$

$\sin (90 - 4A) = \sin(A - 20)$

$90 - 4A = A - 20$

$110 = 5A$

$A = 22^{\circ}$

$\displaystyle \cos ^2\ 10^{\circ}-\sin^2 80+\tan ^2 45^{\circ}$

$\displaystyle \cos ^2\ 10^{\circ}-\sin^2 80+\tan ^2 45^{\circ}$
=

$\displaystyle \cos ^2\ (90 - 80)^{\circ}-\sin^2 80+\tan ^2 45^{\circ}$
=

$\displaystyle \sin ^2\ 80^{\circ}-\sin^2 80+\tan ^2 45^{\circ}$
=

$\tan ^2 45^{\circ}$
=

$(1)^2$
=

$1$

State true or false

Given $A= 60^{0}$ and $B= 30^{0}$ , then
$\sin \left ( A+B \right )= \sin A\cos B+\cos A\sin B$
If true enter 1 else 0

$A= 60^{\circ}$ and

$B = 30^{\circ}$

Now,

$\sin \left ( A+B \right )= \sin A\cos B+\cos A\sin B$

L.H.S:

$\sin (A + B) = \sin (60^0 + 30^0) = \sin 90^0 = 1$
R.H.S:

$\sin A \cos B + \cos A \sin B$
=

$\sin 60^0 \cos 30^0 + \cos 60^0 \sin 30^0$
=

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

$\frac{3}{4} + \frac{1}{4}$
=

$1$

Find the value of: $\displaystyle \sqrt{\frac{1-\sin ^{2}60^{0}}{1-\cos ^{2}60^{0}}}$ is $\displaystyle \frac{1}{\sqrt{m}}$ , m is

$\displaystyle \sqrt{\dfrac{1-\sin ^{2}60^{0}}{1-\cos ^{2}60^{0}}}$

$\sqrt{\dfrac{\cos^260^{\circ}}{\sin^2 60^{\circ}}}$

$\dfrac{\cos 60^{\circ}}{\sin 60^{\circ}}$

$\cot 60^{\circ}$

$\dfrac{1}{\sqrt{3}} = \dfrac{1}{\sqrt{m}}$

thus,

$\sqrt{m} = \sqrt{3}$

$m = 3$

$\cos \left ( A-B \right )= \cos A\cos B+\sin A\sin B$
If true or false:

Given,

$A = 60^{\circ}$ and

$B = 30^{\circ}$

Now,

$\cos \left ( A-B \right )= \cos A\cos B+\sin A\sin B$

L.H.S :

$\cos (A - B) = \cos (60 - 30) = \cos 30 = \frac{\sqrt{3}}{2}$

R.H.S:

$\cos A\cos B+\sin A\sin B = \cos 60 \cos 30 + \sin 60 \sin 30$
=

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

$\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4}$
=

$\frac{\sqrt{3}}{2}$

If $A= 60^{0}$ and $B= 30^{0}$ ;
$\displaystyle \frac{\sin \left ( A-B \right )}{\sin A \sin B}= \cot B-\cot A$
If true enter 1 else 0

Given:

$A = 60^{\circ}, B = 30^{\circ}$

Now,

$\displaystyle \frac{\sin \left ( A-B \right )}{\sin A \sin B}= \cot B-\cot A$

L.H.S:

$\displaystyle \frac{\sin \left ( A-B \right )}{\sin A \sin B}$

=

$\displaystyle \frac{\sin \left ( 60 - 30 \right )}{\sin 60 \sin 30}$

=

$\displaystyle \frac{\sin 30}{\sin 60 \sin 30}$

=

$\frac{1}{\sin 60}$

=

$\frac{1}{\frac{\sqrt{3}}{2}}$

=

$\frac{2}{\sqrt{3}}$

R.H.S :

$\cot B - \cot A$
=

$\cot 30 - \cot 60$
=

$\sqrt{3} - \frac{1}{\sqrt{3}}$
=

$\frac{3 - 1}{\sqrt{3}}$
=

$\frac{2}{\sqrt{3}}$

Thus,

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

Refer to the triangle given in figure.
$\sec C$ is $\displaystyle \frac{3\sqrt{2}}{m}$ , m is

Given,

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

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

$AD = 3$

$BD = 4$

$BC = 12$

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,

$\sec C = \frac{H}{P} = \frac{12}{8 \sqrt{2}} = \frac{3}{2\sqrt{2}}$

Consider the following figure:

$\sin B$ is $\displaystyle \frac{12}{m}$ , m is

$\triangle ABC$ ,

$AD \perp BC$

$AB = 13$ ,

$BD = 5$ ,

$DC =16$

Now, In

$\triangle ABD$ ,

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

$13^2 = AD^2 + 5^2$

$AD^2 = 144$

$AD = 12$

Now, in

$\triangle ADC$ ,

$AC^2 = AD^2 + CD^2$

$AC^2 = 12^2 + 16^2$

$AC^2 = 144 + 256$

$AC = 20$ cm

Now,

$\sin B = \frac{P}{H} = \frac{AD}{AB} = \frac{12}{13}$

$\displaystyle 3-\sin ^2 48^{\circ}+\cos ^2 42^{\circ}$

$\displaystyle 3-\sin ^2 48^{\circ}+\cos ^2 42^{\circ}$
=

$\displaystyle 3-\sin ^2 (90 - 42)^{\circ}+\cos ^2 42^{\circ}$
=

$\displaystyle 3-\cos ^2 42^{\circ}+\cos ^2 42^{\circ}$
=

$3$

Consider the following figure:

$\tan C$ is $\displaystyle \frac{m}{4}$ , m is

$\triangle ABC$ ,

$AD \perp BC$

$AB = 13$ ,

$BD = 5$ ,

$DC =16$

Now, In

$\triangle ABD$ ,

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

$13^2 = AD^2 + 5^2$

$AD^2 = 144$

$AD = 12$

Now, in

$\triangle ADC$ ,

$AC^2 = AD^2 + CD^2$

$AC^2 = 12^2 + 16^2$

$AC^2 = 144 + 256$

$AC = 20$ cm

Now,

$\tan C = \frac{P}{B} = \frac{AD}{CD} = \frac{12}{16} = \frac{3}{4}$

Refer to the triangle given in the figure.

If $\cos A$ is equal to $\displaystyle \frac{m}{5}$ , find $m$ .

Given,

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

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

$AD = 3$

$BD = 4$

$BC = 12$

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,

$\cos A = \dfrac{B}{H} = \dfrac{AD}{AB} = \dfrac{3}{5}$

Therefore,

$m$ is

$3$ .

If $\cos A$ is $\displaystyle \frac{3}{m}$ , then m is

Given:

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

$AB = 3$ and

$BC = 4$
In

$\triangle ABC$ ,

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

$3^2 + 4^2 = AC^2$

$AC^2 = 25$

$AC = 5$ cm

$cos A = \frac{AB}{AC} = \frac{3}{5}$

From the following figure, find the values of:

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

$\triangle ABC$ ,

$AD \perp BC$

$AB = 13$ ,

$BD = 5$ ,

$DC =16$

Now, In

$\triangle ABD$ ,

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

$13^2 = AD^2 + 5^2$

$AD^2 = 144$

$AD = 12$

Now, in

$\triangle ADC$ ,

$AC^2 = AD^2 + CD^2$

$AC^2 = 12^2 + 16^2$

$AC^2 = 144 + 256$

$AC = 20$ cm

Now,

$\sec^2 B - \tan^2 B = (\frac{H}{B})^2 - (\frac{P}{B})^2 = (\frac{AB}{BD})^2 - (\frac{AD}{BD})^2$
=

$(\frac{13}{5})^2 - (\frac{12}{5})^2$
=

$\frac{169}{25} - \frac{144}{25}$
=

$\frac{25}{25}$
=

$1$

If $\sin C$ is $\displaystyle \frac{1}{m}$ , then $m$ is:

Given,

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

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

$AD = 3$

$BD = 4$

$BC = 12$

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,

$\sin C = \frac{P}{H} = \frac{BD}{DC} = \frac{4}{12} = \frac{1}{3}$

Refer to the triangle given in figure.
If $cosec\: A$ is $\displaystyle \cfrac{5}{m}$ , then $m$ is:

Given,

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

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

$AD = 3$

$BD = 4$

$BC = 12$

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,

$cosec A = \frac{H}{P} = \frac{AB}{BD} = \frac{5}{4} = 1 \frac{1}{4}$

Also find the value of : $\sin B\cdot \cos C\, +\, \cos B\cdot \sin C$ .

Given:

$\angle A = 90^{\circ}, AB = 8, BC = 17$

By Pythagoras Theorem,

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

$8^2 + AC^2 = 17^2$

$AC^2 = 289 - 64$

$AC^2 = 225$

$AC = 15$ cm

$sin B cos C + cos B sin C$
= $(\cfrac{P}{H}) (\cfrac{B}{H}) + (\cfrac{B}{H}) (\cfrac{P}{H})$
= $(\cfrac{AC}{BC})(\cfrac{AC}{BC}) + (\cfrac{AB}{BC}) (\cfrac{AB}{BC})$
= $(\cfrac{15}{17})(\cfrac{15}{17}) + (\cfrac{8}{17}) (\cfrac{8}{17})$
= $\cfrac{225 + 64}{289}$
= $1$

From the figure, $\sin x^{0}$ is $\displaystyle \frac{\sqrt{m}}{2}$ . The value of $m$ is:

Let the given triangle be ABC, where

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

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

By Pythagoras Theorem,

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

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

$y^2 = 3$

$y = \sqrt{3}$

$\sin x^{\circ} = \frac{BC}{AC} = \frac{\sqrt{3}}{2}$

In triangle $ABC$ , $AD$ is perpendicular to $BC$ . $\sin B= 0.8$ , $BD = 9$ cm and $\tan C= 1$ . Find the length of $AB$

$\triangle ABC$ ,

$AD \perp BC$ ,

$\sin B = 0.8$ ,

$\tan C = 1$ ,

$BD = 9$

In

$\triangle ABD$ ,

$\sin B = 0.8$

$\cos^2 B = 1 - \sin^2 B = 1 - (0.8)^2$

$\cos B = (0.6)^2$

$\cos B = 0.6$

$\cos B = \frac{BD}{AB}$

$AB = \frac{BD}{0.6}$

$AB = \frac{9}{0.6} = 15$
Also, using Pythagoras theorem,

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

$15^2 = 9^2 + AD^2$

$AD = 144$

$AD = 12$ cm

Now, In

$\triangle ACD$

$\tan C = 1 = \frac{P}{B} = \frac{AD}{CD}$

$AD = CD = 12$ cm
Using Pythagoras theorem,

$AD^2 + CD^2 = AC^2$

$12^2 + 12^2 = AC^2$

$AC = 12\sqrt{2}$ cm

In triangle $ABC$ , $AB= AC= 15$ cm and $BC= 18$ cm. Find :

m if $\sin C$ is $\displaystyle \dfrac{4}{m}$ ,

$AB = AC = 15$ ,

$BC = 18$

Using cosine rule,

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

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

$\cos C = \frac{9}{15}$

$\cos C = \frac{3}{5}$

$\cos C = \frac{B}{H} = \frac{3}{5}$

Using Pythagoras Theorem,

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

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

$P = 4$

$\sin C = \frac{P}{H} = \frac{4}{5}$

Find the length of $AD$ . If $\angle ABC = 60^{\circ},\ \angle DBC = 45^{\circ}$ and $BC = 40\ cm.$

$\triangle ABC$ ,

$\tan60^o = \dfrac{AC}{BC}$

$\Rightarrow \sqrt3= \dfrac{AC}{40}$

$\Rightarrow AC = 4 0\sqrt{3}\ cm$

$\triangle DBC$ ,

$\tan 45^o = \dfrac{DC}{BC}$

$\Rightarrow 1 = \dfrac{DC}{4}$

$\Rightarrow DC = 40\ cm$

Now,

$AD = AC - CD$

$\Rightarrow AD = 40\sqrt{3} - 40$

$\Rightarrow AD = 29.28\ cm$

From the following figure, find the value of:

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

$\triangle ABC$ ,

$AD \perp BC$

$AB = 13$ ,

$BD = 5$ ,

$DC =16$

Now, In

$\triangle ABD$ ,

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

$13^2 = AD^2 + 5^2$

$AD^2 = 144$

$AD = 12$

Now, in

$\triangle ADC$ ,

$AC^2 = AD^2 + CD^2$

$AC^2 = 12^2 + 16^2$

$AC^2 = 144 + 256$

$AC = 20$ cm

Now,

$\sin^2 C + \cos^2 C = (\dfrac{P}{H})^2 + (\dfrac{B}{H})^2 = (\dfrac{AD}{AC})^2 + (\dfrac{BD}{AC})^2$
=

$(\dfrac{12}{20})^2 + (\dfrac{16}{20})^2$
=

$\dfrac{144}{400} + \dfrac{256}{400}$
=

$\dfrac{400}{400}$
=

$1$

$\cos ^{2}A\, +\, \sin ^{2}A$

Given,

$\triangle ABC$ ,

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

$AB = BC = a$
By Pythagoras Theorem,

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

$AC^2 = a^2 + a^2$

$AC = a\sqrt{2}$

Now,

$\cos^2 A + \sin^2 A = (\frac{B}{H})^2 + (\frac{P}{H})^2 = (\frac{a}{a\sqrt{2}})^2 + (\frac{a}{a\sqrt{2}})^2$
=

$\frac{1}{2} + \frac{1}{2}$
=

$1$

Find $AB$ , if:

$\triangle ABC$

$\tan 45 = \frac{AB}{BC}$

$AB = BC = x$

In

$\triangle ABD$ ,

$\tan 30 = \dfrac{AB}{BD}$

$BD = AB \sqrt{3}$

$BC + CD = AB \sqrt{3}$

$x + 20 = x\sqrt{3}$

$x = \dfrac{20}{\sqrt{3} - 1}$

$x = \dfrac{20}{\sqrt{3} - 1}\times\dfrac{\sqrt3+1}{\sqrt3+1}$

$x = 10(\sqrt3+1)=27.32$

Thus,

$AB = BC = 27.32$ cm

Find $AC$

Given:

$AB = 12$ ,

$\angle ACB = 30^{\circ}$

Now,

$\sin 30 = \frac{P}{H} = \frac{AB}{AC}$

$\frac{1}{2} = \frac{12}{AC}$

$AC = 24$

Find $AB$

$\triangle ABC$

$\tan 60 = \frac{AB}{BC}$

$AB = \sqrt{3} BC = \sqrt{3}x$
Let

$BC = x, AB = \sqrt{3} x$

In

$\triangle ABD$ ,

$\tan 45 = \frac{AB}{BD}$

$BD = AB$

$BC + CD = AB$

$x + 20 = x\sqrt{3}$

$x = \frac{20}{\sqrt{3} - 1}$

$x = 27.32$

Thus,

$BC = 27.32$ cm and

$AB = 27.32 \sqrt{3} = 47.32$

The value of $BC$ in the closest integer (in cm.) is:

Given:

$AB = 12$ ,

$\angle ACB = 30^{\circ}$

Now,

$\tan 30 = \frac{P}{B} = \frac{AB}{BC}$

$\frac{1}{\sqrt{3}} = \frac{12}{BC}$

$BC = 12 \sqrt{3}$

$BC = 20.78$ cm

$Find \space BC \space and \space AB$

$\textbf{Step 1: Find relation between AB and BC}$

$\text{In }$

$\triangle ABD, CD=20 \space cm$

$\text{Let}$

$BC = x$

$\therefore \tan C = \dfrac{\text{Opposite Side}}{\text{Adjacent Side}}$

$\Rightarrow \tan 60 = \dfrac{AB}{BC}$

$\Rightarrow \sqrt3 = \dfrac{AB}{BC}$

$\Rightarrow AB = \sqrt{3} BC$

$\therefore AB = \sqrt{3} x$

$\textbf{Step 2: Find Value of AB and BC}$

$\text{In }$

$\triangle ABD,$

$\Rightarrow \tan 30 = \dfrac{AB}{BD}$

$\therefore BD = AB \sqrt{3}$

$\Rightarrow BC + CD =AB \sqrt{3}$

$\Rightarrow x + 20 = x\sqrt{3} (\sqrt{3})$

$\Rightarrow x + 20 = 3x$

$\Rightarrow 2x = 20$

$\therefore BC= x = 10$

$\because AB = \sqrt{3} x$

$\Rightarrow AB=\sqrt3 \times 10$

$\therefore AB=17.32 \space cm$

$\textbf{Thus,BC = 10, AB = 17.32 cm}$

Find $PQ$ , if $\displaystyle AB = 150\ m, \angle P = 30^{\circ}$ and $\displaystyle \angle Q = 45^{\circ}$

$\triangle APB$ ,

$\tan 30 = \frac{AB}{PB}$

$PB = 150 \sqrt{3}$

In

$\triangle AQB$

$\tan 45 = \frac{AB}{BQ}$

$1 = \frac{150}{BQ}$

$BQ = 150$

Thus,

$PQ = PB - BQ$

$PQ = 150\sqrt{3} - 150$

$PQ = 109.8$

cos A $=\displaystyle \dfrac{3}{m}$ ,

Given:

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

$AB = 3$ and

$BC = 4$

$\triangle ABC$ ,

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

$3^2 + 4^2 = AC^2$

$AC^2 = 25$

$AC = 5\ cm$

$\cos A = \dfrac{AB}{AC} = \dfrac{3}{5}$

$\text{but,}\ \cos A = \dfrac{3}{m}$

by comparing we get,

$m=5$

If $\tan x^{\circ} = \dfrac {5}{12}, \tan y^{\circ} = \dfrac {3}{4}$ and $AB = 48\ m$ ; find the length of $CD$ .

$\tan x = \frac{5}{12}$

$\tan y = \frac{3}{4}$

$AB = 48$
Let

$CD = m$

Then,

$\tan x = \frac{CD}{AC} =\frac{5}{12}$

$AC = \frac{12 m}{5}$

$\tan y = \frac{CD}{BC} =\frac{3}{4}$

$BC = \frac{4 m}{3}$
Given,

$AB = AC - BC$

$48 = \frac{12 m}{5} - \frac{4m}{3}$

$48 = \frac{36 m - 20 m}{15}$

$16 m = 48\times 15$

$m = 45$
Thus,

$CD = 45$ m

From the given figure, find the value of $\sin^{2}B\, +\, \cos^{2}B$

Given:

$\angle A = 90^{\circ}, AB = 8, AC = 17$

By Pythagoras Theorem,

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

$8^2 + AC^2 = 17^2$

$AC^2 = 289 - 64$

$AC^2 = 225$

$AC = 15\text{ cm}$

$\displaystyle \sin^2 B + \cos^2 B = \left(\frac{AC}{BC}\right)^2 + \left(\frac{AB}{BC}\right)^2$

$=\displaystyle \left(\frac{15}{17}\right)^2 + \left(\frac{8}{17}\right)^2$

$=\displaystyle \frac{225 + 64}{289}$

$=1$

In the following figure; angle $\displaystyle B=90^{\circ},\angle ADB=30^{\circ}$ and $\displaystyle \angle ACB=60^{\circ}$ .
If $\displaystyle CD = 40\ m$ , then $AB$ is 34.64 m
If true then enter $1$ and if false then enter $0$

$\triangle ABC$

$\tan 60 = \frac{AB}{BC}$

$AB = \sqrt{3} BC$
Let

$BC = x$ ,

$AB = \sqrt{3} x$

In

$\triangle ABD$ ,

$\tan 30 = \frac{AB}{BD}$

$BD = AB \sqrt{3}$

$BC + CD = AB \sqrt{3}$

$x + 40 = x\sqrt{3} (\sqrt{3})$

$x + 40 = 3x$

$x = 20$

Thus,

$AB = 20 \sqrt{3}= 34.64$ cm

cos A is $\displaystyle \frac{12}{m}$ , m is

$\tan A = \frac{5}{12}$

$\tan A = \frac{P}{B} = \frac{5}{12}$

$P = 5, B = 12$

Now,

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

$H^2 = 5^2 + 12^2$

$H^2 = 169$

$H = 13$

$\cos A = \frac{B}{H} = \frac{12}{13}$

$\therefore m=13$

sin A is $\displaystyle \frac{m}{13}$ , m is

$\tan A = \frac{5}{12}$

$\tan A = \frac{P}{B} = \frac{5}{12}$

$P = 5, B = 12$

Now,

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

$H^2 = 5^2 + 12^2$

$H^2 = 169$

$H = 13$

$\cos A = \frac{P}{H} = \frac{5}{13}$

From the given figure, find the value of sin B.cos C + cos B.sin C

Given:

$\angle A = 30^{\circ}, AB = 8, AC = 17$

By Pythagoras Theorem,

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

$8^2 + AC^2 = 17^2$

$AC^2 = 289 - 64$

$AC^2 = 225$

$AC = 15$ cm

$sin B cos C + cos B sin C$

$\displaystyle (\frac{Opposite \ Side}{Hypotenuse}) (\frac{Adjacent \ Side}{Hypotenuse}) + (\frac{Opposite \ Side}{Hypotenuse}) (\frac{Adjacent \ Side}{Hypotenuse})$

$\displaystyle (\frac{AC}{BC})(\frac{AC}{BC}) + (\frac{AB}{BC}) (\frac{AC}{BC})$

$\displaystyle (\frac{15}{17})(\frac{15}{17}) + (\frac{8}{17}) (\frac{8}{17})$

$\displaystyle \frac{225 + 64}{289}$

$1$

sin B is $\displaystyle \frac{m}{5}$ , m is

$AB = AC = 5$ ,

$BC = 8$

Using cosine rule,

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

$5^2 = 8^2 + 5^2 - 2 (5) (8) \cos B$

$\cos B = \dfrac{8}{10}$

$\cos B = \dfrac{4}{5}$

$\cos B = \dfrac{B}{H} = \dfrac{4}{5}$

Now, Using Pythagoras Theorem,

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

$5^2 = P^2 + 4^2$

$P = 3$

Thus,

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

In triangle $ABC$ , $AD$ is perpendicular to $BC$ .If $\sin B = 0.8, BD = 9 cm$ and $\tan C = 1$ . Find the length of $AB$

Given :

$\triangle ABC$ ,

$AD \perp BC$ ,

$\sin B = 0.8$ ,

$\tan C = 1$ ,

$BD = 9$

In

$\triangle ABD$ ,

$\sin B = 0.8$

$\cos^2 B = 1 - \sin^2 B = 1 - (0.8)^2$

$\cos B = (0.6)^2$

$\cos B = 0.6$

$\cos B = \dfrac{BD}{AB}$

$AB = \dfrac{BD}{0.6}$

$AB = \dfrac{9}{0.6} = 15$
Also, using Pythagoras theorem,

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

$15^2 = 9^2 + AD^2$

$AD = 144$

$AD = 12$ cm

Now, In

$\triangle ACD$

$\tan C = 1 = \dfrac{P}{B} = \dfrac{AD}{CD}$

$AD = CD = 12$ cm
Using Pythagoras theorem,

$AD^2 + CD^2 = AC^2$

$12^2 + 12^2 = AC^2$

$AC = 12\sqrt{2}$ cm

If $\sin\, 30^{\circ}\, \cos\, 30^{\circ}$ is $\displaystyle \dfrac{\sqrt{3}}{m}$ , then $m$ is:

$\sin30^{\circ} \cos 30^{\circ} = \dfrac{1}{2} \times \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}}{4}$

If $3 \tan A - 5 \cos B =$ $\sqrt{3}$ and $B =$ $90^{\circ}$ , find the value of $A$ in degrees.

$3 \tan A - 5 \cos B = \sqrt{3}$

$3 \tan A - 5 \cos 90^0 = \sqrt{3}$

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

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

$\tan A = \tan 30^0$

$A = 30^{\circ}$

AB is 17.32 cm
If true then enter $1$ and if false then enter $0$

$\triangle ABC$

$\tan 60 = \frac{AB}{BC}$

$AB = \sqrt{3} BC$
Let

$BC = x$ ,

$AB = \sqrt{3} x$

In

$\triangle ABD$ ,

$\tan 30 = \frac{AB}{BD}$

$BD = AB \sqrt{3}$

$BC + CD = AB \sqrt{3}$

$x + 20 = x\sqrt{3} (\sqrt{3})$

$x + 20 = 3x$

$x = 10$

Thus, BC = 10, AB = 17.32 cm

Find PQ, if AB = 150 m, $\angle P\, =\, 30^{\circ}\, and\, \angle Q\, =\, 45^{\circ}$

$\triangle APB$ ,

$\tan 30 = \frac{AB}{PB}$

$PB = 150 \sqrt{3}$

In

$\triangle AQB$

$\tan 45 = \frac{AB}{BQ}$

$1 = \frac{150}{BQ}$

$BQ = 150$

Thus,

$PQ = PB - BQ$

$PQ = 150\sqrt{3} - 150$

$PQ = 109.8$

Ab is 47.32 cm
If true then enter $1$ and if false then enter $0$

$\triangle ABC$

$\tan 60 = \frac{AB}{BC}$

$AB = \sqrt{3} BC = x$
Let

$BC = x, AB = \sqrt{3} x$

In

$\triangle ABD$ ,

$\tan 45 = \frac{AB}{BD}$

$BD = AB$

$BC + CD = AB$

$x + 20 = x\sqrt{3}$

$x = \frac{20}{\sqrt{3} - 1}$

$x = 27.32$

Thus,

$AB = 27.32$ cm and

$BC = 27.32 \sqrt{3} = 47.32$

$\cos^{2}\, 10^{\circ}\, -\, \sin^{2}\, 80^{\circ}\, +\, \tan^{2}\, 45^{\circ}$

$\cos^{2}\, 10^{\circ}\, -\, \sin^{2}\, 80^{\circ}\, +\, \tan^{2}\, 45^{\circ}$
=

$\cos^{2}\, (90 - 80)^{\circ}\, -\, \sin^{2}\, 80^{\circ}\, +\, \tan^{2}\, 45^{\circ}$
=

$\sin^{2}\, 80^{\circ}\, -\, \sin^{2}\, 80^{\circ}\, +\,\tan^{2}\, 45^{\circ}$
=

$\tan^2 {45}^{\circ}$
=

$1^2$
=

$1$

AB is 27.32 cm
If true then enter $1$ and if false then enter $0$

$\triangle ABC$

$\tan 45 = \dfrac{AB}{BC}$

$AB = BC = x$

In

$\triangle ABD$ ,

$\tan 30 = \dfrac{AB}{BD}$

$BD = AB \sqrt{3}$

$BC + CD = AB \sqrt{3}$

$x + 20 = x\sqrt{3}$

$x = \dfrac{20}{\sqrt{3} - 1}$

$x = 27.32$

Thus,

$AB = BC = 27.32$ cm

$sin\,x^{\circ}$ is $\displaystyle\,\frac{\sqrt{3}}{m}$
value of m is

Given:

$\Delta\,ABC,\,\angle\,B\,=\,90^{\circ}$ , AB = y units, BC =

$\sqrt{3}$ units, AC = 2 units and angle A =

$x^{\circ}$ ,

Using Pythagoras theorem,

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

$2^2 = y^2 + (\sqrt{3})^2$

$y = 1$

Now,

$\sin x^{\circ} = \frac{P}{H} = \frac{BC}{AC} = \frac{\sqrt{3}}{2}$

The value of $x$ is $m\sqrt{2}$ , so $m$ is equal to:

Given:

$\triangle ABC$ ,

$BC = AB = 5$

$\tan \theta = \frac{P}{B} = \frac{BC}{AB}$

$\tan \theta = \frac{5}{5} = 1$

$\tan \theta = \tan 45$

$\theta = 45^{\circ}$

Now,

$\sin \theta = \frac{P}{H}$

$\sin 45 = \frac{AB}{AC}$

$\frac{1}{\sqrt{2}} = \frac{5}{AC}$

$AC = 5 \sqrt{2}$

Use $cos\,x^{\circ}$ to find the value of y.

Given:

$\Delta\,ABC,\,\angle\,B\,=\,90^{\circ}$ , AB = y units, BC =

$\sqrt{3}$ units, AC = 2 units and angle A =

$x^{\circ}$ ,

Now,

$\sin x^{\circ} = \frac{P}{H} = \frac{BC}{AC} = \frac{\sqrt{3}}{2}$

$\sin x^{\circ} = \frac{\sqrt{3}}{2}$

$\sin x = \sin 60$

$x = 60^{\circ}$

Now,

$\cos 60 = \frac{B}{H}$

$\frac{1}{2} = \frac{AB}{AC}$

$\frac{1}{2} = \frac{y}{2}$

$y = 1$

If $\sec A \tan B+\tan A \sec B=91$ , then the value of $(\sec A \sec B+\tan A \tan B)^2$ is equal to

Given,

$\sec A \tan B+\tan A \sec B=91$

Squaring both sides, we get

$\Rightarrow \sec^2 A\tan^2 B+\tan^2 A\sec^2 B+2\tan A\tan B\sec A\sec B=8281$

$\Rightarrow 2\tan A\tan B\sec A\sec B=8281-\sec^2 A\tan^2 B-\tan^2 A\sec^2 B......(1)$

Consider,

$(sec A sec B+tan A tan B)^2$

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

$=\sec^2 A\sec^2 B+\tan^2 A\tan^2 B+8281-\sec^2 A\tan^2 B-\tan^2 A\sec^2 B$

$(by (1))$

$=\sec^2 A\sec^2 B-\sec^2 A\tan^2 B-\tan^2 A\sec^2 B+\tan^2 A\tan^2 B+8281$

$=\sec^2 A(\sec^2 B-\tan^2 B)-\tan^2 A(\sec^2 B-\tan^2 B)+8281$

$=\sec^2 A -\tan^{2} A+8281$

$=1+8281$

$=8282$

Hence, the value is

$8282$

If $\displaystyle \cos \theta =\frac{3}{4}$ then the value of $\displaystyle \sqrt{\frac{\sec \theta -cosec \theta }{\sec \theta +cosec \theta }}$ is ____

$\displaystyle \cos \theta =\dfrac{3}{4}$

$\cos \theta =\dfrac { 3 }{ 4 } =\dfrac { b }{ h }$

$p=\sqrt { { h }^{ 2 }-{ b }^{ 2 } } =\sqrt { { 4 }^{ 2 }-3^{ 2 } }$

$p=\sqrt { { 4 }^{ 2 }-3^{ 2 } } =\sqrt { 16-9 } =\sqrt { 7 }$

$cosec\theta =\dfrac { h }{ p } =\dfrac { 4 }{ \sqrt { 7 } }$

$sec\theta =\dfrac { h }{ b } =\dfrac { 4 }{ 3 }$

$\sqrt { \dfrac { \sec \theta -cosec\theta }{ \sec \theta +cosec\theta } } =\sqrt { \dfrac { \dfrac { 4 }{ 3 } -\dfrac { 4 }{ \sqrt { 7 } } }{ \dfrac { 4 }{ 3 } +\dfrac { 4 }{ \sqrt { 7 } } } }$

$\sqrt { \dfrac { \sec \theta -cosec\theta }{ \sec \theta +cosec\theta } } =\sqrt { \dfrac { \dfrac { 4(\sqrt { 7 } -3) }{ 3\sqrt { 7 } } }{ \dfrac { 4(\sqrt { 7 } +3) }{ 3\sqrt { 7 } } } }$

Find the value of $\displaystyle \frac{2\sin 68^{\circ}}{\cos 22^{\circ}}-\frac{2\cot 15^{\circ}}{5\tan 75^{\circ}}-\frac{3\tan 45^{\circ}\tan 20^{\circ}\tan 40^{\circ}\tan 50^{\circ}\tan 70^{\circ}}{5}$ .

$\Rightarrow$

$\displaystyle \frac{2\sin 68^{\circ}}{\cos 22^{\circ}}-\frac{2\cot 15^{\circ}}{5\tan 75^{\circ}}-\frac{3\tan 45^{\circ}\tan 20^{\circ}\tan 40^{\circ}\tan 50^{\circ}\tan 70^{\circ}}{5}$

$\displaystyle =\frac{2\sin (90^{\circ}-22^{\circ})}{\cos 22^{\circ}}-\frac{2\cot (90^{\circ}-75^{\circ})}{5\tan 75^{\circ}}-\frac{3(1)(\tan 20^{\circ}\tan 70^{\circ})(\tan 40^{\circ}\tan 50^{\circ})}{5}$

$=\displaystyle \frac{2\cos 22^{\circ}}{\cos 22^{\circ}}-\frac{2\tan 75^{\circ}}{5\tan 75^{\circ}}-\frac{3[\tan (90^{\circ}-70^{\circ})\tan 70^{\circ}][\tan (90^{\circ}-50^{\circ})\tan 50^{\circ}]}{5}$

$\displaystyle =2-\frac{2}{3}-\frac{3}{5}(\cot 70^{\circ}\tan 70^{\circ})(\cot 50^{\circ}\tan 50^{\circ})$

$\displaystyle [\therefore \tan (90^{\circ}-\theta )=\cot \theta ,\cot (90^{\circ}-\theta )=\tan \theta \: \&\: \sin (90^{\circ}-\theta )=\cos \theta ]$

$\displaystyle =2-\frac{2}{5}-\frac{3}{5}=2-1=1$

Find the value of $\displaystyle \frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}-2$ .

$\displaystyle \frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}-2$

$\displaystyle =\frac{\cot (90^{\circ}-36^{\circ})}{\tan 36^{\circ}}+\frac{\tan (90^{\circ}-70^{\circ})}{\cot 70^{\circ}}-2$

$\displaystyle =\frac{\tan 36^{\circ}}{\tan 36^{\circ}}+\frac{\cot 70^{\circ}}{\cot 70^{\circ}}-2$ [

$\displaystyle \because \cot (90-\theta )=\tan \theta \: \: and\: \: \tan (90-\theta )=\cot \theta$ ]

$= 1 + 1 - 2 = 0$

Hence, the answer is

$0$ .

The value of $\displaystyle \cos ^{2}5^{0}+\cos ^{2}10^{0}+\cos ^{2}15^{0}+\cos ^{2}20^{0}+\cos ^{2}70^{0}+\cos ^{2}75^{0}+\cos ^{2}80^{0}+\cos ^{2}85^{0}$ is equal to

The value of

$\displaystyle \cos ^{2}5^{0}+\cos ^{2}10^{0}+\cos ^{2}15^{0}+\cos ^{2}20^{0}+\cos ^{2}70^{0}+\cos ^{2}75^{0}+\cos ^{2}80^{0}+\cos ^{2}85^{0}$

$\displaystyle =\left ( \cos ^{2}5^{0}+\sin ^{2}5^{0} \right )+\left ( \cos ^{2}10^{0}+\sin ^{2}10^{0} \right )+\left ( \cos ^{2}15^{0}+\sin ^{2}15^{0} \right )+\left ( \cos ^{2}20^{0}+\sin ^{2}20^{0} \right )$

$=1+1+1+1=4$

$\displaystyle \left [ \because \cos 85^{0}=\cos \left ( 90^{0}-5^{0} \right )=\sin 5^{0}etc \right ]$

Eliminate $\theta$ , $\displaystyle x=a\cos ^{3}\theta\ y=b\sin ^{3}\theta$

$\displaystyle x=a\cos ^{3}\theta \Rightarrow \cos ^{3}\theta =\frac{x}{a}$ or

$\displaystyle \cos \theta =\left ( \frac{x}{a} \right )^{\cfrac{1}{3}}$
Similarly

$\displaystyle \sin \theta =\left ( \frac{y}{b} \right )^{\cfrac{1}{3}}$

$\displaystyle \therefore \sin ^{2}\theta +\cos ^{2}\theta =\left ( \cfrac{x}{a} \right )^{\cfrac{2}{3}}+\left ( \frac{y}{b} \right )^{\cfrac{2}{3}}$

$\displaystyle \because \left ( \frac{x}{a} \right )^{\cfrac{2}{3}}+\left ( \frac{y}{b} \right )^{\cfrac{2}{3}}=1$

Find the relation obtained by eliminating $\displaystyle \theta$ from the equations $x = r$ $\displaystyle \cos \theta +s\sin \theta$ and $y=r$ $\displaystyle \sin \theta -s\cos \theta$

Given,

$x = r$

$\displaystyle \cos \theta +s\sin \theta$

$\displaystyle \Rightarrow x^{2} =\left ( r\cos\theta+s\sin\theta \right )^{2}\\=r^{2}\cos ^{2}\theta+ 2rs \cos \theta. \sin \theta +s^{2} \sin ^{2}\theta$
Also

$y = r \displaystyle \sin \theta-s \cos\theta\\ \Rightarrow y^{2}=r^{2}\sin ^{2}\theta+ s^{2}\cos ^{2}\theta -2rs \sin \theta \cos \theta$

$\displaystyle \Rightarrow x^{2}+y^{2}=r^{2}\left ( \cos ^{2}\theta +\sin ^{2}\theta \right )+s^{2}\left ( \sin ^{2}\theta +\cos ^{2}\theta \right )$

$\displaystyle =r^{2}\left ( 1 \right )+s^{2}\left ( 1 \right )\left [ \because \sin ^{2}\theta +\cos ^{2}\theta =1 \right ]\\=r^{2}+s^{2}$
Hence, the required relation is

$\displaystyle x^{2}+y^{2}=r^{2}+s^{2}$

Show that : (i) $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ} = 1$
(ii) $\cos 38^{\circ} \cos 52^{\circ} - \sin 38^{\circ} \sin 52^{\circ} = 0$ .

$\textbf{Step 1: Apply complementary angle property of trigonometric function and simplify.}$

$(i) \tan 48^{\circ}\tan 23^{\circ}\tan 42^{\circ}\tan 67^{\circ}$

$= \tan 48^{\circ}\tan 23^{\circ}\tan (90^{\circ} - 48^{\circ})\tan (90^{\circ} - 23^{\circ})$

$= \tan 48^{\circ}\tan 23^{\circ}\cot 48^{\circ}\cot 23^{\circ}$

$\boldsymbol{[\because\tan(90^{\circ} - \theta) = \cot\theta]}$

$= (\tan 48^{\circ}.\cot 48^{\circ})(\tan 23^{\circ}\cot 23^{\circ})$

$= 1\times 1$

$\boldsymbol{[\because\tan\theta\times\cot\theta = 1]}$

$= 1$

$\text{(Proved)}$

$(ii) \cos 38^{\circ}\cos 52^{\circ} - \sin 38^{\circ}\sin 52^{\circ}$

$= \cos(38^{\circ} + 52^{\circ})$

$\boldsymbol{[\because\cos(A + B) = \cos A\cos B - \sin A\sin B]}$

$= \cos 90^{\circ}$

$= 0$

$\text{(Proved)}$

$\textbf{Hence, proved.}$

Evaluate:
(i) $\displaystyle \frac { \sin { { 1 }8^{ \circ } } }{ \cos { { 72 }^{ \circ } } }$

(ii) $\displaystyle \frac { \tan { { 26 }^{ \circ } } }{ \cot { { 64 }^{ \circ } } }$

(iii) $\displaystyle \cos { { 48 }^{ \circ } } -\sin { 42^{ \circ } }$

(iv) $\displaystyle \csc \ { 31 }^{ \circ }-\sec { { 59 }^{ \circ } }$

(i)

$\dfrac{\sin18^o}{\cos 72^o}=\dfrac{\sin(90-72)^o}{\cos 72^o}=\dfrac{\cos 72^o}{\cos 72^o}=1$

Hence, the value is

$1$

(ii)

$\dfrac{\tan26^o}{\cot 64^o}=\dfrac{\tan(90-64)^o}{\cot 64^o}=\dfrac{\cot 64^o}{\cot 64^o}=1$

Hence, the value is

$1$

(iii)

$\cos 48^o-\sin 42^o$

$= \cos(90-42)^o-\sin 42^o$

$=\sin 42^o -\sin 42^o$

$=0$

Hence, the value is

$0$

(iv)

$\csc \ 31^o-\sec 59^o$

$=\csc\ (90-59)^o-\sec 59^o$

$=\sec 59^o-\sec 59^o$

$=0$

Hence, the value is

$0$

If $\cos{x} = \dfrac{24}{25}$ , then $\sec{x} =$

We know that

$\sec { x }$ is inverse of

$\cos { x }$ that is

$\sec { x } =\cos ^{ -1 }{ x } =\dfrac { 1 }{ \cos { x } }$ .

Here, it is given that

$\cos { x } =\dfrac { 24 }{ 25 }$ , therefore,

$\sec { x } =\dfrac { 1 }{ \cos { x } } =\dfrac { 1 }{ \dfrac { 24 }{ 25 } } =1\times \dfrac { 25 }{ 24 } =\dfrac { 25 }{ 24 }$

Hence,

$\sec { x } =\dfrac { 25 }{ 24 }$ .

If $\sin { \theta } =\cfrac { 1 }{ 2 }$ and $\cos { \theta } =-\cfrac { \sqrt { 3 } }{ 2 }$ , find the values of the other 4 trignometric functions.

Given $\sin \theta = \dfrac{1}{2}$ and $\cos \theta = -\dfrac{\sqrt3}{2}$
Which implies $\tan \theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{\dfrac 12}{-\dfrac{\sqrt3}2} = -\dfrac{1}{\sqrt3}$
$\cot \theta = \dfrac{1}{\tan\theta} = -\sqrt3$
$\sec\theta = \dfrac{1}{\cos\theta} = -\dfrac{2}{\sqrt3}$
$\text{cosec}\theta = \dfrac 1{\sin\theta} = 2$

If $cosec\ {x} = \dfrac{25}{15}$ , $\sin{x} =$

We know that

$\sin { x }$ is inverse of

$cosec\ { x }$ that is

$\sin { x } =\dfrac { 1 }{ cosec\ { x } }$ .

Here, it is given that

$cosec\ { x } =\dfrac { 25 }{ 15 }$ , therefore,

$\sin { x } =\dfrac { 1 }{ cosec\ { x } } =\dfrac { 1 }{ \dfrac { 25 }{ 15 } } =1\times \dfrac { 15 }{ 25 } =\dfrac { 15 }{ 25 }$

Hence,

$\sin { x } =\dfrac { 15 }{ 25 }$ .

If $\tan{x} = \dfrac{7}{24}$ , $\cot{x} =$

We know that

$\cot { x }$ is inverse of

$\tan { x }$ that is

$\cot { x } =\dfrac { 1 }{ \tan { x } }$ .

Here, it is given that

$\tan { x } =\dfrac { 7 }{ 24 }$ , therefore,

$\cot { x } =\dfrac { 1 }{ \tan { x } } =\dfrac { 1 }{ \dfrac { 7 }{ 24 } } =1\times \dfrac { 24 }{ 7 } =\dfrac { 24 }{ 7 }$

Hence,

$\cot { x } =\dfrac { 24 }{ 7 }$ .

If $\cot{A} = \dfrac{8}{15}$ and $\sin{A} = \dfrac{15}{17}$ then, $\cos{A} =$

We know that

$\cot { x }$ is

$\cos { x }$ over

$\sin { x }$ that is

$\cot { x } =\dfrac { \cos { x } }{ \sin { x } }$ .

Here, it is given that

$\sin { A } =\dfrac { 15 }{ 17 }$ and

$\cot { A } =\dfrac { 8 }{ 15 }$ , therefore,

$\cot { A } =\dfrac { \cos { A } }{ \sin { A } } \\ \Rightarrow \cos { A } =\sin { A } \times \cot { A } =\dfrac { 15 }{ 17 } \times \dfrac { 8 }{ 15 } =\dfrac { 8 }{ 17 }$

Hence,

$\cos { A } =\dfrac { 8 }{ 17 }$ .

If $\sin{x} = \dfrac{3}{5}$ , then $\text{ cosec} \ x = ?$

We know that

$cosec \ { x }$ is inverse of

$\sin { x }$ that is

$\csc { x } =\sin ^{ -1 }{ x } =\dfrac { 1 }{ \sin { x } }$ .

Here, it is given that

$\sin { x } =\dfrac { 3 }{ 5 }$ , therefore,

$cosec \ { x } =\dfrac { 1 }{ \sin { x } } =\dfrac { 1 }{ \dfrac { 3 }{ 5 } } =1\times \dfrac { 5 }{ 3 } =\dfrac { 5 }{ 3 }$

Hence,

$cosec\ { x } =\dfrac { 5 }{ 3 }$ .

What trigonometric ratios of angles from ${0}^{o}$ to ${90}^{o}$ are not defined?

The trigonometric table of all the functions with all values are shown in the above table:

Hence, from the above table we observe that between

$0^o$ to

$90^o$ , the below trigonometric functions are equal to not defined.

$\tan { \dfrac { \pi }{ 2 } } =\csc { { 0 }^{ o } } =\cot { { 0 }^{o } } =\sec { \dfrac { \pi }{ 2 } }=$ Not defined

Find the value of ' $x$ '

In a right

$\triangle KLM,\quad \angle L={ 30 }^{ 0 }$

We know that

$\tan { \theta } =\dfrac { Opposite\quad side }{ Adjacent\quad side } =\dfrac { KM }{ LM }$

Here,

$\theta ={ 30 }^{ 0 }$ ,

$KM=x$ and

$LM=100$ m, therefore,

$\tan { \theta } =\dfrac { KM }{ LM } \\ \Rightarrow \tan { 30^{ 0 } } =\dfrac { x }{ 100 } \\ \Rightarrow \dfrac { 1 }{ \sqrt { 3 } } =\dfrac { x }{ 100 } \quad \quad \quad \quad \quad \quad \left( \because \quad \tan { 30^{ 0 } } =\dfrac { 1 }{ \sqrt { 3 } } \right) \\ \Rightarrow \sqrt { 3 } x=100\\ \Rightarrow x=\dfrac { 100 }{ \sqrt { 3 } } \\ \Rightarrow x=\dfrac { 100\times \sqrt { 3 } }{ \sqrt { 3 } \times \sqrt { 3 } } =\dfrac { 100\times \sqrt { 3 } }{ 3 } =\dfrac { 100\sqrt { 3 } }{ 3 }$

Hence,

$x=\dfrac { 100\sqrt { 3 } }{ 3 }$ m.

Find the value of ' $x$ '

In a right

$\triangle PQR,\quad \angle Q={ 60 }^{ 0 }$

We know that

$\tan { \theta } =\dfrac { Opposite\quad side }{ Adjacent\quad side } =\dfrac { PR }{ PQ }$

Here,

$\theta ={ 60 }^{ 0 }$ ,

$PQ=x$ and

$PR=90$ m, therefore,

$\tan { \theta } =\dfrac { PR }{ PQ } \\ \Rightarrow \tan { 60^{ 0 } } =\dfrac { 90 }{ x } \\ \Rightarrow \sqrt { 3 } =\dfrac { 90 }{ x } \quad \quad \quad \quad \quad \quad \left( \because \quad \tan { 60^{ 0 } } =\sqrt { 3 } \right) \\ \Rightarrow \sqrt { 3 } x=90\\ \Rightarrow x=\dfrac { 90 }{ \sqrt { 3 } } \\ \Rightarrow x=\dfrac { 90\times \sqrt { 3 } }{ \sqrt { 3 } \times \sqrt { 3 } } =\dfrac { 90\times \sqrt { 3 } }{ 3 } =30\sqrt { 3 }$

Hence,

$x=30\sqrt { 3 }$ m.

If $A={ 60 }^{ o },B={ 30 }^{ o }$ then prove that $\cos { \left( A+B \right) } =\cos { A } \cos { B } -\sin { A } \sin { B }$

LHS

$=\cos { \left( A+B \right) } =\cos { \left( { 60 }^{ o }+{ 30 }^{ o } \right) } =\cos { { 90 }^{ o } } =0$
RHS

$=\cos { A } \cos { B } -\sin { A } \sin { B }$

$=\cos { { 60 }^{ o } } \cos { { 30 }^{ o } } -\sin { { 60 }^{ o } } \sin { { 30 }^{ o } }$

$=\dfrac { 1 }{ 2 } \times \dfrac { \sqrt { 3 } }{ 2 } -\dfrac { \sqrt { 3 } }{ 2 } \times \dfrac { 1 }{ 2 } =\dfrac { \sqrt { 3 } }{ 4 } -\dfrac { \sqrt { 3 } }{ 4 } =0$

$\therefore$ LHS

$=$ RHS

What trigonometric ratios of angles from ${0}^{o}$ to ${90}^{o}$ are equal?

The trigonometric table of all the functions with all values are shown in the above table:

Hence, from the above table we observe that between

$0^o$ to

$90^o$ , the below trigonometric functions are equal.

$\sin { \dfrac { \pi }{ 2 } } =\csc { \dfrac { \pi }{ 2 } } =1$

$\cos { { 0 }^{ o } } =\sec { { 0 }^{ o } } =1$

$\sin { \dfrac { \pi }{ 4 } } =\cos { \dfrac { \pi }{ 4 } } =\dfrac { \sqrt { 2 } }{ 2 }$

$\sec { \dfrac { \pi }{ 4 } } =\csc { \dfrac { \pi }{ 4 } } =\sqrt { 2 }$

$\tan { \dfrac { \pi }{ 4 } } =\cot { \dfrac { \pi }{ 4 } } =1$

What trigonometric ratios of angles from ${0}^{\circ}$ to ${90}^{\circ}$ are equal to $0$ ?

The trigonometric table of all the functions with all values are shown in the above table:

Hence, from the given table we observe that between

$0^o$ to

$90^o$ , the below trigonometric functions are equal to

$0$ .

$\sin { { 0 }^{ o } } =\tan { { 0 }^{ o} } =\cos { \cfrac { \pi }{ 2 } } =\cot { \cfrac { \pi }{ 2 } } =0$

If $A={ 60 }^{ o }$ and $B={ 30 }^{ o }$ , then prove that $\tan { \left( A-B \right) } =\dfrac { \tan { A } -\tan { B } }{ 1+\tan { A } \tan { B } }$

In the given identity

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

$A ={ 60 }^{ 0 }$ and

$B ={ 30 }^{ 0 }$ as shown below:

$\tan { (A-B) } =\dfrac { \tan { A } -\tan { B } }{ 1+\tan { A } \tan { B } } \\ \Rightarrow \tan { ({ 60 }^{ 0 }-{ 30 }^{ 0 }) } =\dfrac { \tan { { 60 }^{ 0 } } -\tan { { 30 }^{ 0 } } }{ 1+\tan { { 60 }^{ 0 } } \tan { { 30 }^{ 0 } } } \\ \Rightarrow \tan { { 30 }^{ 0 } } =\dfrac { \tan { { 60 }^{ 0 } } -\tan { { 30 }^{ 0 } } }{ 1+\tan { { 60 }^{ 0 } } \tan { { 30 }^{ 0 } } }$

We know that the values of the trignometric functions

$\tan { { 30 }^{ 0 } } =\dfrac { 1 }{ \sqrt { 3 } }$ and

$\tan { { 60 }^{ 0 } } =\sqrt { 3 }$ .

The value of left hand side(LHS) is:

$\tan { { 30 }^{ 0 } } =\dfrac { 1 }{ \sqrt { 3 } } ..........(1)$

Let us now find the value of the right hand side (RHS) that is

$\dfrac { \tan { { 60 }^{ 0 } } -\tan { { 30 }^{ 0 } } }{ 1+\tan { { 60 }^{ 0 } } \tan { { 30 }^{ 0 } } }$ as shown below:

$\dfrac { \tan { { 60 }^{ 0 } } -\tan { { 30 }^{ 0 } } }{ 1+\tan { { 60 }^{ 0 } } \tan { { 30 }^{ 0 } } } =\dfrac { \sqrt { 3 } -\dfrac { 1 }{ \sqrt { 3 } } }{ 1+\left( \sqrt { 3 } \times \dfrac { 1 }{ \sqrt { 3 } } \right) } =\dfrac { \dfrac { 3-1 }{ \sqrt { 3 } } }{ 1+1 } =\dfrac { \dfrac { 2 }{ \sqrt { 3 } } }{ 2 } =\dfrac { 2 }{ \sqrt { 3 } } \times \dfrac { 1 }{ 2 } =\dfrac { 1 }{ \sqrt { 3 } } \quad ...........(2)$

From equations 1 and 2, we get that

$LHS=RHS$

Hence,

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

$A ={ 60 }^{ 0 }$ and

$B ={ 30 }^{ 0 }$ .

What trigonometric ratios of angles from ${0}^{o}$ to ${90}^{o}$ are equal to $1$ ?

The trignometric table of all the functions with all values are shown in the above table:

Hence, from the above table we observe that between

$0^o$ to

$90^o$ , the below trignometric functions are equal to

$1$ .

$\sin { \dfrac { \pi }{ 2 } } =\cos { { 0 }^{ o } } =\csc { \dfrac { \pi }{ 2 } } =\sec { { 0 }^{ o } } =\tan { \dfrac { \pi }{ 4 } } =\cot { \dfrac { \pi }{ 4 } } =1$

Prove that $\cos { \theta } \cdot \text{cosec} { \theta } =\cot { \theta }$

LHS

$=\cos { \theta } \text{cosec} { \theta }$

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

$\left[ \because \text{cosec} { \theta } =\dfrac { 1 }{ \sin { \theta } } \right]$

$\therefore \dfrac { \cos { \theta } }{ \sin { \theta } } =\cot { \theta } =$ RHS

Find the value of ' $x$ '

In a right

$\triangle XZY,\quad \angle Z={ 45 }^{ 0 }$

We know that

$\cos { \theta } =\dfrac { Adjacent\quad side }{ Hypotenuse } =\dfrac { YZ }{ XZ }$

Here,

$\theta ={ 45 }^{ 0 }$ ,

$YZ=x$ and

$XZ=100$ m, therefore,

$\cos { \theta } =\dfrac { YZ }{ XZ } \\ \Rightarrow \cos { 45^{ 0 } } =\dfrac { x }{ 100 } \\ \Rightarrow \dfrac { 1 }{ \sqrt { 2 } } =\dfrac { x }{ 100 } \quad \quad \quad \quad \quad \quad \left( \because \quad \cos { 45^{ 0 } } =\dfrac { 1 }{ \sqrt { 2 } } \right) \\ \Rightarrow \sqrt { 2 } x=100\\ \Rightarrow x=\dfrac { 100 }{ \sqrt { 2 } } \\ \Rightarrow x=\dfrac { 100\times \sqrt { 2 } }{ \sqrt { 2 } \times \sqrt { 2 } } =\dfrac { 100\times \sqrt { 2 } }{ 2 } =50\sqrt { 2 }$

Hence,

$x=50\sqrt { 2 }$ m.

Prove that $\dfrac { \sin { \theta } }{ 1-\cot { \theta } } +\dfrac { \cos { \theta } }{ 1-\tan { \theta } } =\cos { \theta } +\sin { \theta }$ .

Given :

$\dfrac { \sin { \theta } }{ 1-\cot { \theta } } +\dfrac { \cos { \theta } }{ 1-\tan { \theta } } =\cos { \theta } +\sin { \theta }$
L.H.S.

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

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

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

$=\dfrac { \left( \sin { \theta } +\cos { \theta } \right) \left( \sin { \theta } -\cos { \theta } \right) }{ \sin { \theta } -\cos { \theta } }$

$=\sin { \theta } +\cos { \theta } =$ R.H.S.
L.H.S

$=$ R.H.S.

Evaluate without using trigonometric tables
$\cos^226^\circ+\cos\,64^\circ \sin\,26^\circ+\dfrac{\tan\,36^\circ}{\cot\,54^\circ}$

$\cos ^226^\circ +\cos 64^\circ \sin 26^\circ +\dfrac{\tan 36^\circ }{\cot 54^\circ }=\cos ^226^\circ +\cos (90^\circ -26^\circ )\sin 26^\circ +\dfrac{\tan 36^\circ }{\cot (90^\circ -36^\circ )}$

$=\cos ^226^\circ +\sin 26^\circ \sin 26^\circ +\dfrac{\tan 36^\circ }{\tan 36^\circ }$

$=\cos ^226^\circ +\sin ^226^\circ +1$

$=1+1$

$=2$

Without using tables evaluate $3\cos 80^o.\text{cosec}\, 10^o+2\sin 59^o.\sec 31^o$ .

$3\cos 80^\circ . \text{cosec}\, 10^\circ +2\sin 59^\circ. \sec 31^\circ$

$=3\cos 80^\circ \text{cosec}\, (90^\circ -80^\circ )+2\sin 59^\circ sec(90^\circ -59^\circ )$

$=3\cos 80^\circ \sec 80^\circ +2\sin 59^\circ \text{cosec}\, 59^\circ$

$=3\cos 80^\circ \times \displaystyle\frac{1}{\cos 80^\circ }+2\sin 59^\circ \times \displaystyle\frac{1}{\sin 59^\circ }$

$=3+2=5$ .

If $\cos { 7A } =\sin { \left( A-{ 6 }^{ o } \right) }$ , where $7A$ is an acute angle. Find the value of $A$ .

When angle is acute,

$\cos \theta =\sin (90^\circ-\theta )$

$\cos 7A=\sin (90^\circ-7A)$

$\Rightarrow \sin ({ 90 }^{ \circ }-7A)=\sin (A-{ 6 }^{ \circ })$

$\Rightarrow { 90 }^{ \circ }-7A=A-{ 6 }^{ \circ }$

$\Rightarrow 8A=96^{ \circ }\\ \Rightarrow A=12^\circ$

Without using trigonometric tables evaluate :
$\dfrac { \sin { { 63 }^\circ } }{ \cos { { 27 }^\circ } } +\dfrac { \cos { { 39 }^\circ } }{ \sin { { 51 }^\circ } } -\sin { { 23 }^\circ } \cdot \sec { { 67 }^\circ } +\csc ^{ 2 }{ { 30 }^\circ }$

$\dfrac { \sin { { 63 }^\circ } }{ \cos { { 27 }^\circ } } +\dfrac { \cos { { 39 }^\circ } }{ \sin { { 51 }^\circ } } -\sin { { 23 }^\circ } \cdot \sec { { 67 }^\circ } +\csc ^{ 2 }{ { 30 }^\circ }$

$=\dfrac { \cos { \left( { 90 }^\circ-{ 65 }^\circ \right) } }{ \cos { { 25 }^\circ } } +\dfrac { \sin { \left( { 90 }^\circ-{ 32 }^\circ \right) } }{ \sin { { 58 }^\circ } } -\cos { \left( { 90 }^\circ-{ 28 }^\circ \right) } \times \dfrac { 1 }{ \cos { { 62 }^\circ } } +{ \left( 2 \right) }^{ 2 }$

$=\dfrac { \cos { { 25 }^\circ } }{ \cos { { 25 }^\circ } } +\dfrac { \sin { { 58 }^\circ } }{ \sin { { 58 }^\circ } } -\cos { { 62 }^\circ } \times \dfrac { 1 }{ \cos { { 62 }^\circ } } +4$

$=1+1-1+4$

$=5$

If $\cos{\theta}=\dfrac{1}{2}$ , then find the value of acute angle $\theta$ .

Given,

$\cos { \theta } =\dfrac { 1 }{ 2 }$ ....(i)
But

$\cos { { 60 }^{ o } } =\dfrac { 1 }{ 2 }$ ....(ii)
From equation (i) and (ii), we have

$\cos { \theta } =\cos { { 60 }^{ o } }$

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

Prove that: $\dfrac {\cos A}{1 + \sin A} + \tan A = \sec A$

L.H.S.

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

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

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

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

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

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

$= \sec A =$ R.H.S.

If $\theta = 60^{\circ}$ , find the value of $\cos \theta$

Given,

$\theta = 60^{\circ}$ (given)

$\therefore \cos \theta = \cos (60^{\circ})$

$= \cos 60^{\circ}$

$= \dfrac {1}{2}$

Prove that $\displaystyle \frac{\tan^2\theta}{(\sec\,\theta-1)^2}=\frac{1+\cos\,\theta}{1-\cos\,\theta}$

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

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

$=\cfrac{\sec\theta+1}{\sec\theta-1}$

$=\cfrac{\cfrac1{\cos\theta}+1}{\cfrac1{\cos\theta}-1}$
Multiply numerator and denominator with

$cos{\theta}$ , we get

$=\cfrac{1+\cos\theta}{1-\cos\theta}$ , Hence proved

Without using trigonometric tables, evaluate $\sin^{2} 34^{\circ} + \sin^{2} 56^{\circ} + 2\tan 18^{\circ} \tan 72^{\circ} - \cot^{2} 30^{\circ}$

Given:

$\sin^{2} 34^{\circ} + \sin^{2} 56^{\circ} + 2\tan 18^{\circ} \tan 72^{\circ} - \cot^{2} 30^{\circ}$

$= \sin^{2} 34^{\circ} + \sin^{2} (90^{\circ} - 34^{\circ}) + 2\tan 18^{\circ} \tan (90^{\circ} - 18^{\circ}) - \cot^{2} 30^{\circ}$

$= \sin^{2} 34^{\circ} + \cos^{2} 34^{\circ} + 2\tan 18^{\circ} \cot 18^{\circ} - (\sqrt {3})^{2}$

$= 1 + 2\tan 18^{\circ} \times \dfrac {1}{\tan 18^{\circ}} - 3$

$= 1 + 2 - 3$

$= 0$

Verify the following equalities.
$\cos 60^{\circ} = 1 - 2\sin^{2} 30^{\circ} = 2\cos^{2} 30^{\circ} - 1$

$\cos { 60° } =\dfrac { 1 }{ 2 }$

$\Rightarrow 1-2\sin ^{ 2 }{ 30° } =1-2{ \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }=1-\dfrac { 1 }{ 2 } =\dfrac { 1 }{ 2 }$

$\Rightarrow 2\cos ^{ 2 }{ 30° } -1=2{ \left( \dfrac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }=\dfrac { 3 }{ 2 } -1=\dfrac { 1 }{ 2 }$

$\therefore \cos { 60° } =1-2\sin ^{ 2 }{ 30° } =2\cos ^{ 2 }{ 30° } -1$

Hence, the answer is verify.

Verify the following equalities.
$\sin^{2}30^{\circ} + \cos^{2} 30^{\circ} = 1$

$\sin ^{ 2 }{ 30° } +\cos ^{ 2 }{ 30° } =1$

$\Rightarrow L.H.S=\sin ^{ 2 }{ 30° } +\cos ^{ 2 }{ 30° } ={ \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }+{ \left( \dfrac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }=\dfrac { 1 }{ 4 } +\dfrac { 3 }{ 4 }$

$=\dfrac { 4 }{ 4 }$

$=1$

$\Rightarrow R.H.S=1=L.H.S$

Hence, the answer is verify.

Verify the following equalities.
$\cos 90^{\circ} = 1 - 2\sin^{2} 45^{\circ} = 2\cos^{2} 45^{\circ} - 1$

$\cos { 90° } =0,1-2\sin ^{ 2 }{ 45° } =1-2{ \left( \dfrac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }=1-1=0$

$\Rightarrow 2\cos ^{ 2 }{ 45° } -1=2\times { \left( \dfrac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }-1=1-1=0$

$\therefore \cos { 90° } =1-2\sin ^{ 2 }{ 45° } =2\cos ^{ 2 }{ 45° }$

Hence, the answer is verified.

Prove that $\sqrt {\dfrac {1 + \cos \theta}{1 - \cos \theta}} = cosec \theta + \cot \theta$

$\textbf{Step1: Simplify the given expression}$

$\text{L.H.S}$

$=\sqrt{\cfrac{1+\cos\theta}{1-\cos\theta}}$

$=\sqrt{\cfrac{1+\cos\theta}{1-\cos\theta}\times\cfrac{1+\cos\theta}{1+\cos\theta}}$

$\textbf{[Multiplying numerator and denominator with}$

$\boldsymbol{1+\cos\theta \ }]$

$=\sqrt{\cfrac{(1+\cos\theta)^2}{1-\cos^2\theta}}$

$=\sqrt{\cfrac{(1+\cos\theta)^2}{\sin^2\theta}}$

$=\cfrac{1+\cos\theta}{\sin\theta}$

$=\cfrac{1}{\sin\theta}+\cfrac{\cos\theta}{\sin\theta}$

$=\text{cosec}\,\theta+\cot\theta$

$=R.H.S.$

$\textbf{Hence proved.}$

Prove that
$\dfrac {\sin (90^{\circ} - \theta)}{1 + \sin \theta} + \dfrac {\cos \theta}{1 - \cos (90^{\circ} - \theta)} = 2\sec \theta$

$LHS = \dfrac {\sin (90^{\circ} - \theta)}{1 + \sin \theta} + \dfrac {\cos \theta}{1 - \cos (90^{\circ} - \theta)}$

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

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

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

$= \dfrac {2\cos \theta}{\cos^{2}\theta}$

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

$= 2\sec \theta$

$LHS = RHS$

Verify the following equalities.
$1 + \tan^{2} 45^{\circ} = \sec^{2} 45^{\circ}$

$1+\tan ^{ 2 }{ 45° } =\sec ^{ 2 }{ 45° }$

$\Rightarrow L.H.S=1+\tan ^{ 2 }{ 45° } =1+1=2$

$\Rightarrow R.H.S=\sec ^{ 2 }{ 45° } ={ \left( \sqrt { 2 } \right) }^{ 2 }=2$

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

Hence, the answer is verify.

Verify the following equalities.
$\tan^{2} 60^{\circ} - 2\tan^{2} 45^{\circ} - \cot^{2} 30^{\circ} + 2\sin^{2} 30^{\circ} + \dfrac {3}{4} cosec^{2} 45^{\circ} = 0$

$\tan ^{ 2 }{ 60° } -2\tan ^{ 2 }{ 45° } -\cot ^{ 2 }{ 30° } +2\sin ^{ 2 }{ 30° } +\dfrac { 3 }{ 4 } cosec^{ 2 }45°$

$={ \left( \sqrt { 3 } \right) }^{ 2 }-2{ \left( 1 \right) }^{ 2 }-{ \left( \sqrt { 3 } \right) }^{ 2 }+2{ \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }+{ \dfrac { 3 }{ 4 } \left( \sqrt { 2 } \right) }^{ 2 }$

$=3-2-3+2\times \dfrac { 1 }{ 4 } +\dfrac { 3 }{ 4 } \times 2$

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

$=-2+2$

$=0.$

Hence, the answer is

$0.$

Verify the given equalities.
$\sin 30^{\circ} \cos 60^{\circ} + \cos 30^{\circ} \sin 60^{\circ} = \sin 90^{\circ}$

$\sin { 30°\cos { 60° } } +\cos { 30°\sin { 60° } } =\sin { 90° }$

$\Rightarrow L.H.S=\sin { 30°\cos { 60° } } +\cos { 30°\sin { 60° } }$

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

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

$=\dfrac { 4 }{ 4 }$

$=1$

$\Rightarrow R.H.S=\sin { 90° } =1$

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

Hence, answer verify.

Verify the following equalities.
$\dfrac {1 - \tan^{2} 60^{\circ}}{1 + \tan^{2} 60^{\circ}} = 2\cos^{2} 60^{\circ} - 1$

$\dfrac { 1-\tan ^{ 2 }{ 60° } }{ 1+\tan ^{ 2 }{ 60° } } =\dfrac { 1-{ \left( \sqrt { 3 } \right) }^{ 2 } }{ 1+{ \left( \sqrt { 3 } \right) }^{ 2 } } =\dfrac { 1-3 }{ 1+3 } =-\dfrac { 1 }{ 2 }$

$\Rightarrow 2\cos ^{ 2 }{ 60° } -1=2\times { \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }-1=\dfrac { 1 }{ 2 } -1=-\dfrac { 1 }{ 2 }$

$\therefore \dfrac { 1-\tan ^{ 2 }{ 60° } }{ 1+\tan ^{ 2 }{ 60° } } =2\cos ^{ 2 }{ 60° } -1$

Hence, the answer is verify.

Verify the following equalities.
$4\cot^{2} 45^{\circ} - \sec^{2} 60^{\circ} + \sin^{2} 60^{\circ} + \cos^{2} 60^{\circ} = 1$

$4\cot ^{ 2 }{ 45° } -\sec ^{ 2 }{ 60° } +\sin ^{ 2 }{ 60° } +\cos ^{ 2 }{ 60° }$

$=4{ \left( 1 \right) }^{ 2 }-{ \left( 2 \right) }^{ 2 }+{ \left( \dfrac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }+{ \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }$

$=-2+\dfrac { 1 }{ 4 } +\dfrac { 3 }{ 4 }$

$=-2+1$

$=-1.$

Hence, the answer is

$-1.$

Simplify $\dfrac {\cos 80^{\circ}}{\sin 10^{\circ}} + \cos 59^{\circ} cosec 31^{\circ}$

$\dfrac { \cos { 80° } }{ \sin { 10° } } +\cos { 59° } cosec31°=\dfrac { \cos { \left( 90°-10° \right) } }{ \sin { 10° } } +\cos { \left( 90°-31° \right) } cosec31°$

$=\dfrac { \sin { 10° } }{ \sin { 10° } } +\sin { 31° } cosec31°$

$=1+1$

$=2.$

Hence, the answer is

$2.$

Verify the following equalities.
$\dfrac {\sec 30^{\circ} + \tan 30^{\circ}}{\sec 30^{\circ} - \tan 30^{\circ}} = \dfrac {1 + \sin 30^{\circ}}{1 - \sin 30^{\circ}}$

$\dfrac { \sec { 30° } +\tan { 30° } }{ \sec { 30° } -\tan { 30° } } =\dfrac { 1+\sin { 30° } }{ 1-\sin { 30° } }$

$\Rightarrow L.H.S=\dfrac { \sec { 30° } +\tan { 30° } }{ \sec { 30° } -\tan { 30° } } =\dfrac { \dfrac { 1 }{ \cos { 30° } } +\dfrac { \sin { 30° } }{ \cos 30° } }{ \dfrac { 1 }{ \cos { 30° } } -\dfrac { \sin { 30° } }{ \cos 30° } }$

$=\dfrac { \left( 1+\sin { 30° } \right) }{ \cos { 30° } } \times \dfrac { \cos { 30° } }{ 1-\sin { 30° } }$

$=\dfrac { 1+\sin { 30° } }{ 1-\sin { 30° } } =R.H.S$

Hence, the answer is verify.

Verify the following equalities.
$\dfrac {\cos 60^{\circ}}{1 + \sin 60^{\circ}} = \dfrac {1}{\sec 60^{\circ} + \tan 60^{\circ}}$

$\dfrac { \cos { 60° } }{ 1+\sin { 60° } } =\dfrac { \dfrac { 1 }{ 2 } }{ 1+\dfrac { \sqrt { 3 } }{ 2 } } =\dfrac { 1 }{ 2 } \times \dfrac { 2 }{ 2+\sqrt { 3 } } =\dfrac { 1 }{ 2+\sqrt { 3 } }$

$\Rightarrow \dfrac { 1 }{ \sec { 60° } +\tan { 60° } } =\dfrac { 1 }{ 2+\sqrt { 3 } } =\dfrac { \cos { 60° } }{ 1+\sin { 60° } }$

Hence, the answer is verify.

Show that: $\tan 35^{\circ} \tan 60^{\circ} \tan 55^{\circ} \tan 30^{\circ} = 1$

To show:

$\tan 35^{\circ} \tan 60^{\circ} \tan 55^{\circ} \tan 30^{\circ} = 1$

We have,

$\tan 35^{\circ} = \tan (90^{\circ} - 55^{\circ}) = \cot 55^{\circ}$
and

$\tan 60^{\circ} = \tan (90^{\circ} - 30^{\circ}) = \cot 30^{\circ}$

$\therefore \tan 35^{\circ} \tan 60^{\circ} \tan 55^{\circ} \tan 30^{\circ}$

$= \cot 55^{\circ} \cot 30^{\circ} \tan 55^{\circ} \tan 30^{\circ}$

$= \dfrac {1}{\tan 55^{\circ}}\times \dfrac {1}{\tan 30^{\circ}}\times \tan 55^{\circ} \times \tan 30^{\circ}$

$= 1$

Hence, the required value is

$1$ .

Prove that: $\sec^2\theta =1+\tan^2\theta$ .

Let

$\Delta ABP$ is a right angle triangle, where

$\angle B=90^o$ .

Let

$\angle PAB=\theta$ .

so,

$\tan\theta=\dfrac{PB}{AB}.....(1)$

and,

$\sec\theta=\dfrac{PA}{AB}......(2)$

By applying Pythagoras theorem, in

$\Delta ABP$ ,

$PA^2=PB^2+AB^2$
Dividing both sides by

$AB^2$

$\Rightarrow \displaystyle\frac{PA^2}{AB^2}=\frac{PB^2}{AB^2}+\frac{AB^2}{AB^2}$

$\Rightarrow \left(\displaystyle\frac{PA}{AB}\right)^2=\left(\displaystyle\frac{PB}{AB}\right)^2+\left(\displaystyle\frac{AB}{AB}\right)^2$ [from

$(1)$ and

$(2)$ ]

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

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

$Hence, proved$ .

Length of two sides of a triangle are $6$ centimetres and $5$ centimetres. If their included angle is ${50}^{o}$ . What is the area of the triangle?
Also find the length of its third side.
$\left( \sin { { 50 }^{ o } } =0.77,\cos { { 50 }^{ o } } =0.64,\tan { { 50 }^{ o } } =1.19 \right)$

Consider the following figure of triangle:
Area of triangle

$=\cfrac { 1 }{ 2 } \times BC\times AD=\cfrac { 1 }{ 2 } \times 6\times AD=3\times AD$
From right-angled

$\triangle{ABD}$ ,

$\sin { { 50 }^{ o } } =\cfrac { AD }{ AB }$

$AD=AB\times \sin { { 50 }^{ o } } =5\times 0.77=3.85cm$
Thus, area of triangle

$3\times 3.85=11.55{cm}^{2}$
Now,

$\cos { { 50 }^{ o } } =\cfrac { BD }{ AB } \quad$

$\Rightarrow BD=AB\times \cos { { 50 }^{ o } } =5\times 0.64=3.2cm$
Thus,

$DC=6-3.2=2.8cm$

$\Rightarrow AC=\sqrt { { (2.8) }^{ 2 }+{ (3.85) }^{ 2 } } =\sqrt { 7.84+14.82 } =\sqrt { 22.66 } cm\quad \quad$

In $\Delta ABC,$ AB = 8 cm, AC = 5 cm and $m \angle A = 50^o$ . Then
(a) What is the length of the perpendicular from C to AB?
(b) Find the length of BC.
$[sin 50^o = 0.7660, cos 50^o = 0.6428, tan 50^o = 1.1918]$

Let CD be perpendicular from C to AB.
(a) In

$\Delta ACD$ ,

$sin 50^o = \dfrac{CD}{AC} \Rightarrow 0.7660 = \dfrac{CD}{5}$

$\Rightarrow CD = 0.7660 \times 5 = 3.83 cm$
(b)

$cos 50^o = \dfrac{AD}{AC} \Rightarrow 0.6428 = \dfrac{AD}{5}$

$\Rightarrow AD = 0.6428 \times 5 = 3.214$
DB = AB - AD = (8 - 3.214) cm = 4.786 cm
In right triangle CDB,

$BC^2 = CD^2 + DB^2$

$= (3.83)^2 + (4.786)^2$

$= 14.67 + 22.91$

$= 37.58$

$\Rightarrow BC = \sqrt{37.58} cm$

Without using trigonometric tables, evaluate the following:
$2\left (\dfrac {\cos 58^{\circ}}{\sin 32^{\circ}}\right ) - \sqrt {3}\left (\dfrac {\cos 38^{\circ} \text{cosec }52^{\circ}}{\tan 15^{\circ} \tan 60^{\circ} \tan 75^{\circ}}\right )$ .

Let

$I=2\left( \dfrac { \cos{ 58 }^{ 0 } }{ \sin{ 32 }^{ 0 } } \right) -\sqrt { 3 } \left( \dfrac { \cos{ 38 }^{ 0 }\text{cosec }{ 52 }^{ 0 } }{ \tan{ 15 }^{ 0 }\tan{ 60 }^{ 0 }\tan{ 75 }^{ 0 } } \right)$

We have

$\tan(90-\theta )=\cot\theta$ and

$\cos(90-\theta ) =\sin\theta$

$=2\left( \dfrac { \cos(90-32) }{ \sin{ 32 }^{ 0 } } \right) -\sqrt { 3 } \left( \dfrac { \cos(90-52)\text{cosec }{ 52 }^{ 0 } }{ \tan{ 15 }^{ 0 }\tan{ 60 }^{ 0 }\tan(90-15) } \right)$

$=2\left( \dfrac { \sin{ 32 }^{ 0 } }{ \sin{ 32 }^{ 0 } } \right) -\sqrt { 3 } \left( \dfrac { \sin{ 52 }^{ 0 }\text{cosec }{ 52 }^{ 0 } }{ \tan{ 15 }^{ 0 }\times \sqrt { 3 } \times cot{ 15 }^{ 0 } } \right)$

We have

$\sin\theta \times \text{cosec}\theta =1$ and

$\tan\theta \times cot\theta =1$

Using this we evaluate the above expression as

$I =2-\sqrt { 3 } \left( \dfrac { 1 }{ \sqrt { 3 } } \right) =2-1=1$

Evaluate : $\displaystyle \, \frac{cosec 13^{\circ}}{\sec 77^{\circ}} \, - \, \frac{\cot 20^{\circ}}{\tan 70^{\circ}}$

${ \dfrac { { cosec\;13° } }{ \sec { 77° } } }-\dfrac { \cot { 20° } }{ \tan { 70° } } =\dfrac { cosec\left( 90°-77° \right) }{ \sec { 77° } } -\dfrac { \cot { \left( 90°-70° \right) } }{ \tan { 70° } }$

$=\dfrac { \sec { 77° } }{ \sec { 77° } } -\dfrac { \tan { 70° } }{ \tan { 70° } }$ {

$cosec(90-A) = \sec A$ and

$\cot(90-A) = \tan A$ }

$=1-1$

$=0$

Hence, the answer is

$0.$

Calculate
$\displaystyle\, \frac{\sin \, \alpha}{\sin^3 \, \alpha \, + \, \cos^3 \, \alpha}$ if $\tan$ $\alpha$ =2.

For such kind of question we have to just simplify the question as much possible, so first we take common

$({ \cos { \alpha } })^{ 3 }$ from denominator and then we got

$\displaystyle \dfrac { \tan { \alpha } \times { (\sec { \alpha } })^{ 2 } }{ 1+({ \tan { \alpha } })^{ 3 } }$

and using trignometric relation we convert sec into tan after that

$\displaystyle\dfrac { \tan { \alpha } \times \left( 1+{ (\tan { \alpha }) }^{ 2 } \right) }{ 1+({ \tan { \alpha } })^{ 3 } }$

and as given the value of

$\tan { \alpha }$ put this and solve simple calculation we got the value of expression is equal to

$\dfrac { 10 }{ 9 }$

If $\cos\theta +\sin\theta =\sqrt{2}\cos\theta$ , show that $\cos\theta -\sin\theta=\sqrt{2}\sin\theta$

$\because \cos \theta + \sin \theta = \sqrt{2} \cos \theta$ ... (i)

Let

$\cos \theta - \sin \theta = k$ ...(ii)

squaring and adding the two equations , we get

$(\cos \theta + \sin \theta)^2 + (\cos \theta - \sin \theta)^2 = (\sqrt{2} \cos \theta)^2 + k^2$

$\Rightarrow \cos^2 \theta + 2 \cos \theta \sin \theta + \sin^2 \theta += \cos^2 \theta - 2 \cos \theta \sin \theta + \sin^2 \theta = 2\cos^2 \theta + k^2$

$\Rightarrow 2 - 2 \cos^2 \theta = k^2$

$\Rightarrow k^2 = 2 \sin^2 \theta$

$\Rightarrow k = \sqrt{2} \sin \theta$

$\therefore \cos \theta - \sin \theta = \sqrt{2} \sin \theta$

Show that:
$\left( 1+\cot { A } +\tan { A } \right) \left( \sin { A } -\cos { A } \right) =\cfrac { \sec { A } }{\text{cosec} ^{ 2 }{ A } } -\cfrac {\text{cosec }{ A } }{ \sec ^{ 2 }{ A } } =\sin { A } \tan { A } -\cot { A } \cos { A }$

$(1 + \cot A + \tan A)(\sin A - \cos A)$

$= \sin A + \sin A \cot A + \sin A \tan A- \cos A -\cos A \cot A - \cos A \tan A$

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

$= \sin A + \cos A + \sin A \tan A - \cos A - \cot A \cos A- \sin A$

$= \sin A \tan A - \cot A \cos A$ .......

$(1)$

Now,

$\sin A \tan A - \cot A \cos A$

$= \sin A \times \dfrac{\sin A}{\cos A} - \dfrac{\cos A}{\sin A} \times \cos A$

$= \sin^2 A \sec A - \cos^2 A \text{ cosec }A$

$= \dfrac{1}{\text{cosec}^2 A} \times \sec A - \dfrac{1}{\sec^2 A} \times \text{cosec }A$

$= \dfrac{\sec A}{\text{cosec}^2 A} - \dfrac{\text{cosec } A}{\sec^2 A}$ ......

$(2)$

From

$(1)$ and

$(2)$ ,

$(1 + \cot A + \tan A)(\sin A - \cos A)$

$= \dfrac{\sec A}{\text{cosec}^2 A} - \dfrac{\text{cosec } A}{\sec^2 A}$

$= \sin A \tan A - \cot A \cos A$

Prove that $\dfrac{\tan^2A}{1 + \tan^2A} + \dfrac{\cot^2A}{1 +\cot^2A} = 1$

$LHS =\dfrac{\tan^2A}{1 + \tan^2A} + \dfrac{\cot^2A}{1 +\cot^2A}$

Using the identites:

$1+\tan^2 A = \sec^2 A$ and

$1+ \cot^2 A = \csc^2 A$ :

$=\dfrac {\tan ^{2}A} {\sec ^{2}A} +\dfrac {\cot ^{2}A} {\csc ^{2}A}$

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

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

$=1$

$=RHS$

Hence proved.

Prove that $\dfrac{\tan A \, + \, \tan B}{\cot A \, + \, \cot B} \, = \, \tan A \, \tan B$

To Prove:

$\dfrac {\tan A+\tan B}{\tan A +\cot B}=\tan A. \tan B$

$L.H.S$

$=\dfrac {\tan A+\tan B}{\dfrac {1}{\tan A}+\dfrac {1}{\tan B}}=\dfrac {\tan A+\tan B}{\dfrac {\tan B+\tan A}{\tan A.\tan B}}\quad \left [\because \ \tan A=\dfrac {1}{\cos A}\right]$

$=\tan A.\tan B=R.H.S$

Prove the following statements.
$\dfrac {\cot A + \tan B}{\cot B + \tan A} = \cot A \tan B$ .

LHS

$=\dfrac { cotA+tanB }{ cotB+tanA }$

$=\dfrac { \left( \dfrac {cosA}{sinA} \right) +\left( \dfrac {sinB}{cosB} \right) }{ \left( \dfrac {cosB}{sinB} \right) +\left( \dfrac {sinA}{cosA} \right) }$

$=\left( \dfrac { cosAcosB+sinAsinB }{ cosBcosA+sinAsinB } \right) \times \dfrac { sinBcosA }{ sinAcosB }$

$=cotAtanB$

Evaluate the following : sin $45^{\circ}$ sin $30^{\circ}$ + cos $45^{\circ}$ cos $30^{\circ}$

Given

$\sin 45^{\circ} \sin 30^{\circ} + \cos 45^{\circ}\cos 30^{\circ}$

$=\left(\dfrac{1}{\sqrt{2}}\right)\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{\sqrt{2}}\right)\left(\dfrac{\sqrt{3}}{2}\right)$

$\left[\because \sin 45^\circ=\cos 45^\circ=\dfrac{1}{\sqrt2},\cos30^\circ=\dfrac{\sqrt3}{2},\sin 30^\circ=\dfrac{1}{2}\right]$

$=\left(\dfrac{1}{2\sqrt{2}}\right)+\left(\dfrac{\sqrt{3}}{2\sqrt{2}}\right)$

$=\left(\dfrac{1+\sqrt{3}}{2\sqrt{2}}\right)$

Prove that $\tan^2A + \cot^2A = \sec^2A.cosec^2A \, - \, 2$

$LHS=\tan^2A + \cot^2A$

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

$=\dfrac {\sin ^{4} A +\cos ^{4}A} {\sin ^{2}A \cos ^{2}A}$

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

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

$=\dfrac1{cos^2A}.\dfrac1{sin^2A}-2$

$=\sec ^{2}A.cosec ^{2}A - 2$

$=RHS$

Hence proved

Prove that $\dfrac{\cos^2 \theta}{\sin \theta} - cosec \theta + \sin \theta = 0$ .

LHS

$=\dfrac{\cos^2 \theta}{\sin \theta} - co\sec \theta + \sin \theta$

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

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

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

$=0$

$=RHS$

Hence proved

Evaluate $\sin 60^{\circ} \cos 30^{\circ} + \cos 60^{\circ} \sin 30^{\circ}$ .

Given

$\sin{60}^{\circ} \cos{30}^{\circ}+\cos{60}^{\circ} \sin{30}^{\circ}$

$=\left(\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{\sqrt{3}}{2}\right)+\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)$

$\left[\because \cos30^\circ=\sin 60^\circ=\dfrac{\sqrt3}{2},\sin 30^\circ=\cos 60^\circ=\dfrac{1}{2}\right]$

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

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

$=\dfrac{4}{4}$

$=1$

Evaluate the following :
cos $60^{\circ}$ cos $45^{\circ}$ - sin $60^{\circ}$ sin $45^{\circ}$

Given

$\cos60^{\circ}\cos45^{\circ}- \sin60^{\circ}\sin45^{\circ}$

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

$\left[\because \sin 45^\circ=\cos 45^\circ=\dfrac{1}{\sqrt2},\sin 60^\circ=\dfrac{\sqrt3}{2},\cos 60^\circ=\dfrac{1}{2}\right]$

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

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

$\dfrac{(1 + \tan^2\theta) \cot \theta}{co\sec^2 \theta} = \tan \theta$

Identity :

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

$LHS=\dfrac{(1 + \tan^2\theta) \cot \theta}{cosec^2 \theta}$

$=\dfrac {\sec ^{2}\theta \cot \theta} {cosec ^{2}\theta}$

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

$=\tan \theta$

$=RHS$

Hence proved

Prove that $\dfrac{\sin 70^{\circ}}{\cos 20^{\circ}} + \dfrac{\text{cosec} 20^{\circ}}{\sec 70^{\circ}} - 2 \cos 70^{\circ} \text{cosec} 20^{\circ} = 0$

To prove that,

$\dfrac{\sin 70^{\circ}}{\cos 20^{\circ}} + \dfrac{cosec 20^{\circ}}{\sec 70^{\circ}} - 2 \cos 70^{\circ} cosec 20^{\circ} = 0$

We know that,

$\cos(90-\theta)=\sin\theta$

$\sec(90-\theta)=cosec\theta$

$cosec(\theta)=\dfrac{1}{\sin\theta}$

$\dfrac{\sin 70^{\circ}}{\cos (90-70)^{\circ}} + \dfrac{\csc 20^{\circ}}{\sec (90-20)^{\circ}} - 2 \ \dfrac{\cos 70^{\circ}}{\sin 20^{\circ}} = 0$

$\dfrac{\sin 70^{\circ}}{\cos (90-70)^{\circ}} + \dfrac{\csc 20^{\circ}}{\sec (90-20)^{\circ}} - 2 \ \dfrac{\cos 70^{\circ}}{\sin (90-70^{\circ})} = 0$

$\dfrac{\sin 70^{\circ}}{\sin70^{\circ}} + \dfrac{\csc 20^{\circ}}{\csc20^{\circ}} - 2 \ \dfrac{\cos 70^{\circ}}{\cos70^{\circ}} = 0$

$1 + 1 - 2 (1) = 0$

$2 - 2 = 0$

$0 = 0$

Hence proved

Express the following in terms of trigonometric ratios of angles lying between $0^\circ$ and $45^\circ$ :
$\cot 85^\circ + \cos 75^\circ$

We know that,

$\cot(90^\circ-\theta)=\tan\theta$ and

$\cos(90^\circ-\theta)=\sin\theta$

$\cot85^\circ=\cot(90^\circ-5^\circ)=\tan5^\circ$

$\cos75^\circ=\cos(90^\circ-15^\circ)=\sin15^\circ$

$\therefore\cot 85^\circ + \cos 75^\circ=\tan5^\circ+\sin15^\circ$

If $A = 30^{\circ}$ and $B = 60^{\circ}$ , verify that
(i) $\sin(A + B) = \sin A \cos B + \cos A \sin B$
(ii) $\cos(A + B) = \cos A \cos B - \sin A \sin B$

Given,

$A = 30^{\circ}$ and

$B = 60^{\circ}$

$\Rightarrow \sin 30^\circ=\cos 60^\circ=\dfrac{1}{2},\cos 30^\circ=\sin 60^\circ=\dfrac{\sqrt3}{2}$

To prove:

(i)

$\sin(A + B) = \sin A \cos B + \cos A \sin B$

$\sin(30^\circ + 60^\circ) = \sin 30^\circ \cos 60^\circ + \cos 30^\circ \sin 60^\circ$

$\sin(90^\circ) = \left(\dfrac{1}{2} \times \dfrac{1}{2}\right) + \left(\dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2}\right)$

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

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

$1=1$

hence verified

(ii)

$\cos(A + B) = \cos A \cos B - \sin A \sin B$

$\cos(30^\circ + 60^\circ) = \cos 30^\circ \cos 60^\circ - \sin 30^\circ \sin 60^\circ$

$\cos(90^\circ) = \left(\dfrac{\sqrt{3}}{2} \times \dfrac{1}{2}\right) - \left(\dfrac{1}{2} \times \dfrac{\sqrt{3}}{2}\right)$

$0 = \dfrac{\sqrt{3}}{4} - \dfrac{\sqrt{3}}{4}$

$0=0$

hence verified

Express the following in terms of trigonometric ratios of angles lying between 0 and 45:
$\text{cosec} 54^0 + \sin 72^0$

We have,

$\csc(90^0-\theta)=\sec\theta$ and

$\sin(90^0-\theta)=\cos\theta$

$\csc54^0=\csc(90^0-36^0)=\sec36^0$

$\sin72^0=\sin(90^0-18^0)=\cos18^0$

$\therefore\csc 54^0 + \sin 72^0=\sec36^0+\cos18^0$

In a rectangle ABCD, $AB = 20$ cm, $\angle$ BAC = $60^{\circ}$ , calculate side $BC$ and diagonals $AC, BD$ .

$\triangle ABC$ , from the above figure

$\tan60^\circ=\dfrac{BC}{AB}=\sqrt3$

$\sqrt3=\dfrac{BC}{20}$

$\therefore \ BC=20\sqrt3 \ cm$

$\cos60^\circ=\dfrac{AB}{AC}=\dfrac{1}{2}$

$\dfrac{1}{2}=\dfrac{20}{AC}$

$\therefore \ AC=40 \ cm$

In a rectangle, the two diagonals are equal.

$\therefore \ AC=BD=40 \ cm$

In a right $\triangle \text{ABC}$ , right angled at $\text{C}$ , if $\angle \text{B} = 60^{\circ}$ and $\text{AB} = 15$ units. Find the remaining angles and sides.

Given:

$\triangle ABC$ is right angled at

$C, \ \angle{B}=60^\circ$

$\angle{C}=90^\circ$

We know that,

In triangle

$ABC$ ,

$\angle{A}+\angle{B}+\angle{C}=180^\circ$

$60^\circ+\angle{A}+90^\circ=180^\circ$

$\angle{A}=30^\circ$

We have,

$\sin{A}=\dfrac{CB}{AB}$

$\sin30^\circ=\dfrac{CB}{15}$

$\dfrac{1}{2}\times 15 = CB$

$CB=7.5 \text{ units}$

From Pythagoras theorem,

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

$15^2=AC^2+\left(\dfrac{15}{2}\right)^2$

$AC^2=225-\left(\dfrac{225}{4}\right)$

$AC^2=225\left(\dfrac{3}{4}\right)$

$AC=\dfrac{15\sqrt{3}}{2}\text{ units}$

Prove the following :
$\dfrac{\cos(90^{\circ} \, - \, \theta) \, \sec \, (90^{\circ} \, - \, \theta) \, \tan \, \theta}{\text{cosec}(90^{\circ} \, - \, \theta) \, \sin(90^{\circ} \, - \, \theta) \, \cot(90^{\circ} \, - \, \theta)} \, + \ \dfrac{\tan(90^{\circ} \, - \, \theta)}{\cot \, \theta} \, = \, 2$

To prove :

$\dfrac{\cos(90^{\circ} \, - \, \theta) \, \sec \, (90^{\circ} \, - \, \theta) \, \tan \, \theta}{\csc(90^{\circ} \, - \, \theta) \, \sin(90^{\circ} \, - \, \theta) \, \cot(90^{\circ} \, - \, \theta)} \, + \ \dfrac{\tan(90^{\circ} \, - \, \theta)}{\cot \, \theta} \, = \, 2$

We know that,

$\cos(90-\theta)=\sin\theta$

$\sin(90-\theta)=\cos\theta$

$\tan(90-\theta)=\cot\theta$

$\cot(90-\theta)=\tan\theta$

$\sec(90-\theta)=\csc\theta$

$\csc(90-\theta)=\sec\theta$

Also,

$\sec(\theta)=\dfrac{1}{\cos(\theta)}$ and $\csc(\theta)=\dfrac{1}{\sin(\theta)}$ ..... (1)

So, rewriting the given equation with these identites, we get

$\dfrac{\sin\theta \, \csc\theta \, \tan\theta}{\sec\theta \, \cos\theta \, \tan\theta} \, + \ \dfrac{\cot\theta}{\cot\theta} \, = \, 2$

$\dfrac{\sin\theta \, \csc\theta}{\sec\theta \, \cos\theta} \, + 1 \, = \, 2$

$\dfrac{\sin\theta \, \cos\theta}{\sin\theta \, \cos\theta} \, + 1 \, = \, 2$ [From (1)]

$1 + 1=2$

$2=2$

Hence proved

If $\Delta ABC$ is a right triangle such that $\angle C = 90^{\circ}, \, \angle A = 45^{\circ}$ and $BC = 7$ units. Find $\angle B, AB$ and $AC$ .

$\triangle ABC$ is

$right \, angled$ at

$C$

$\angle{C}=90^\circ, \angle{A}=45^\circ$

We know that,

In triangle

$ABC$ ,

$\angle{A}+\angle{B}+\angle{C}=180^\circ$

$45^\circ+\angle{B}+90^\circ=180^\circ$

$\angle{B}=45^\circ$

It's a scalene triangle.

$\therefore BC=AC=7 units$

From Pythagoras theorem,

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

$AB^2=7^2+7^2=98$

$AB=7\sqrt2$

Prove that $\sin 48^{\circ} \sec 42^{\circ} + \cos 48^{\circ} \text{cosec} 42^{\circ} = 2$

Given

$\sin 48^{\circ} \sec 42^{\circ} + \cos 48^{\circ} \csc 42^{\circ}$

we know that,

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

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

$\therefore \dfrac{\sin 48^{\circ}}{\cos 42^{\circ}} + \dfrac{\cos 48^{\circ}}{\sin 42^{\circ}}$

We have,

$\cos(90-\theta)=\sin\theta, \ \sin(90-\theta)=\cos\theta$

$\therefore \dfrac{\sin 48^{\circ}}{\cos (90-48)^{\circ}} + \dfrac{\cos 48^{\circ}}{\sin (90-48)^{\circ}}$

$= \dfrac{\sin 48^{\circ}}{\sin 48^{\circ}} + \dfrac{\cos 48^{\circ}}{\cos 48^{\circ}}$

$= 1+1$

$= 2$

$\therefore$

$\sin 48^{\circ} \sec 42^{\circ} + \cos 48^{\circ} \csc 42^{\circ}=2$

hence proved

Prove that: $\tan 20^{\circ} \tan 35^{\circ} \tan 45^{\circ} \tan 55^{\circ} \tan 70^{\circ} = 1$

Given:

$\tan 20^0 \tan 35^0 \tan 45^0 \tan 55^0 \tan 70^0$

this can be written as,

$=\tan (90^0-70^0) \tan (90^0-55^0) \tan 45^0 \tan 55^0 \tan 70^0$

$=\cot 70^0 \cot 55^0 \tan 45^0 \tan 55^0 \tan 70^0$

$[\tan(90-x) = \cot x]$

$=\dfrac{1}{\tan 70^0} \dfrac{1}{\tan 55^0} \tan 45^0 \tan 55^0 \tan 70^0$

$[\cot {x} = \dfrac{1}{\tan x}]$

$=\tan 45^\circ$

$=1$

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

L.H.S

$\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }$

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

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

$\Rightarrow \dfrac{{{\sin }^{2}}\theta }{\cos \theta \left( \sin \theta -\cos \theta \right)}+\dfrac{{{\cos }^{2}}\theta }{\sin \theta \left( \cos \theta -\sin \theta \right)}$

$\Rightarrow \dfrac{{{\sin }^{2}}\theta }{\cos \theta \left( \sin \theta -\cos \theta \right)}-\dfrac{{{\cos }^{2}}\theta }{\sin \theta \left( \sin \theta -\cos \theta \right)}$

$\Rightarrow \dfrac{{{\sin }^{3}}\theta -{{\cos }^{3}}\theta }{\cos \theta \sin \theta \left( \sin \theta -\cos \theta \right)}$

$\Rightarrow \dfrac{\left( \sin \theta -\cos \theta \right)\left( {{\sin }^{2}}+{{\cos }^{2}}\theta +\sin \theta \cos \theta \right)}{\cos \theta \sin \theta \left( \sin \theta -\cos \theta \right)}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\,\,\left[ \because {{x}^{3}}-{{y}^{3}}=\left( x-y \right)\left( {{x}^{2}}+{{y}^{2}}+xy \right) \right]$

$\Rightarrow \dfrac{\left( 1+\sin \theta \cos \theta \right)}{\cos \theta \sin \theta }$

$\Rightarrow 1+\sec \theta \csc \theta$

Hence, proved.

If $3x\sin 60^o.\cos^2 30^o=\dfrac{\tan^2 45^o.\sec 60^o}{cosec 60^o.\sin 45^o}$ , then find the value of $x$ .

$3x\sin 60\cos^230=\dfrac{\tan^245\cdot \sec 60}{cosec 60 \sin 45}$

$3x\times \dfrac{\sqrt{3}}{2}\times \left(\dfrac{\sqrt{3}}{2}\right)^2=\dfrac{(1)^2\times (2)}{\left(\dfrac{2}{\sqrt{3}}\right)\times 1_{\displaystyle /\sqrt{2}}}$

$3x\times \dfrac{\sqrt{\not{3}}}{2}$

$\times \dfrac{3}{4}$

$=\dfrac{\not{2}\sqrt{\not{3}}\times \sqrt{2}}{\not{2}}$

$x=\dfrac{8\sqrt{2}}{9}$ .

If $a \cos \theta - b \sin \theta = x$ and $a \sin \theta + b \cos \theta = y$ then find $\dfrac{a^{2} + b^{2}}{x^{2} + y^{2}}$ .

$x^2=a^2cos^2\theta+b^2sin^2\theta-2abcos\theta sin\theta\\y^2=a^2sin^2\theta+b^2cos^2\theta+2abcos\theta sin\theta$

Now,

$x^2+y^2=a^2cos^2\theta+b^2sin^2\theta-2abcos\theta+a^2sin^2\theta+b^2cos^2\theta+2abcos\theta sin\theta\\=a^2(cos^2\theta+sin^2\theta)+b^2(cos^2\theta+sin^2\theta)=a^2+b^2$

Hence,

$\dfrac{a^2+b^2}{x^2+y^2}=\dfrac{a^2+b^2}{a^2+b^2}=1$

Express each of the following in trigonometric ratios of angles between $0^{\circ} \, and \, 45^{\circ}$ :
$sin 72^{\circ} \, + \, cot 72^{\circ}$

$sin 72^{\circ} \, + \, cot 72^{\circ}$

$= \, sin (90^{\circ} \, - \, 18^{\circ}) \, + \, cot (90^{\circ} \, - \, 18^{\circ})$

$= \, cos 18^{\circ} \, + \, tan 18^{\circ}$

$[ \because \, sin (90^{\circ} \, - \, 18^{\circ}) \, = \, cos 18^{\circ} , cot \, (90^{\circ} \, - \, 18^{\circ}) \, = \, tan 18^{\circ}]$

Prove that :
$sin 35^{\circ} \, sin 55^{\circ} \, - \, cos 35^{\circ} \, cos 55^{\circ} \, = \, 0$

L.H.S. =

$sin 35^{\circ} \, sin 55^{\circ} \, - \, cos 35^{\circ} \, cos 55^{\circ}$

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

$sin (90^{\circ} \, - \, 55^{\circ}) \, sin (90^{\circ} \, - \, 35^{\circ}) \, - \, cos 35^{\circ} \, cos 55^{\circ}$

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

$cos 55^{\circ} \, cos 35^{\circ} \, - \, cos 35^{\circ} \, cos 55^{\circ}$

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

$\Rightarrow \, \,$ L.H.S. = 0 = R.H.S.

Hence proof

The value of $\left( 1+\cos { \dfrac { \pi }{ 8 } } \right) \left( 1+\cos { \dfrac { 3\pi }{ 8 } } \right) \left( 1+\cos { \dfrac { 5\pi }{ 8 } } \right) \left( 1+\cos { \dfrac { 7\pi }{ 8 } } \right) =\dfrac { a }{ 256 }$ . Find a.

$\left( 1+\cos { \dfrac { \pi }{ 8 } } \right) \left( 1+\cos { \dfrac { 3\pi }{ 8 } } \right) \left( 1+\cos { \dfrac { 5\pi }{ 8 } } \right) \left( 1+\cos { \dfrac { 7\pi }{ 8 } } \right) =\dfrac { a }{ 256 }$ .

$\left( 1+\cos { \dfrac { \pi }{ 8 } } \right) \left( 1+\cos { \dfrac { 3\pi }{ 8 } } \right) \left( 1+\cos { (\pi -\dfrac { 3\pi }{ 8 } ) } \right) \left( 1+\cos( { \pi-\dfrac { \pi }{ 8 } }) \right)=\dfrac { a }{ 256 }$ .

$\left( 1+\cos { \dfrac { \pi }{ 8 } } \right) \left( 1+\cos { \dfrac { 3\pi }{ 8 } } \right) \left( 1-\cos { \dfrac { 3\pi }{ 8 } } \right) \left( 1-\cos( { \dfrac { \pi }{ 8 } }) \right)=\dfrac { a }{ 256 }$ .

$\left( 1-\cos^{2} { \dfrac { \pi }{ 8 } } \right)$

$\left( 1-\cos^{2} { \dfrac {3 \pi }{ 8 } } \right)=\dfrac{a}{256}$

$\sin^{2}\dfrac{\pi}{8}\sin^{2}\dfrac{3\pi}{8}=\dfrac{a}{256}$

$\sin^{2}\dfrac{\pi}{8}\sin^{2}(\dfrac{\pi}{4}-\dfrac{\pi}{8})=\dfrac{a}{256}$

$\sin^{2}\dfrac{\pi}{8}\cos^{2}\dfrac{\pi}{8}=\dfrac{a}{256}$

$4\sin^{2}\dfrac{\pi}{8}\cos^{2}\dfrac{\pi}{8}=\dfrac{4a}{256}$

$\sin^{2}\dfrac{\pi}{4}=\dfrac{a}{64}$

(As

$2 \sin A\cos A=\sin 2A$ )

$\dfrac{1}{2}=\dfrac{a}{64}$

$a=32$

Hence the answer is

$a=32$

Find the value of:
$\dfrac{1 + \cos A -\sin A}{1 + \cos A +\sin A} -\sec A + \tan A$

$\dfrac { 1+\cos A-\sin A }{ 1+\cos A+\sin A } -\dfrac { 1 }{ \cos A } +\dfrac { \sin A }{ \cos A } \\ \Rightarrow \dfrac { \left( \cos A \right) \left( 1+\cos A-\sin A \right) -\left( 1+\cos A+\sin A \right) +\sin A\left( 1+\cos A+\sin A \right) }{ \left( \cos A \right) \left( 1+\cos A\sin A \right) } \\ \Rightarrow \dfrac { { \sin }^{ 2 }+{ \cos }^{ 2 }-1 }{ \left( \cos A \right) \left( 1+\cos A\sin A \right) } =0\quad \quad \quad \quad \quad \because \left[ { \sin }^{ 2 }x+{ \cos }^{ 2 }x=1 \right]$

Prove that $\dfrac{1+2\cos^2\,A}{1+3\cot^2\,A}=\sin^2\,A$

To prove:

$\displaystyle{\dfrac { 1+2\cos ^{ 2 }{ A } }{ 1+3\cot ^{ 2 }{ A } } =\sin ^{ 2 }{ A }}$

Let us simplify the left side of the given equation,

$\begin{aligned}{}\frac{{1 + 2{{\cos }^2}A}}{{1 + 3{{\cot }^2}A}} &= \frac{{1 + 2{{\cos }^2}A}}{{1 + 3\dfrac{{{{\cos }^2}A}}{{{{\sin }^2}A}}}}\\ &= {\sin ^2}A\left( {\frac{{1 + 2{{\cos }^2}A}}{{{{\sin }^2}A + 3{{\cos }^2}A}}} \right)\\ &= {\sin ^2}A\left( {\frac{{1 + 2{{\cos }^2}A}}{{1 - {{\cos }^2}A + 3{{\cos }^2}A}}} \right)\\ &= {\sin ^2}A\left( {\frac{{1 + 2{{\cos }^2}A}}{{1 + 2{{\cos }^2}A}}} \right)\\ &= {\sin ^2}A\\&=R.H.S\end{aligned}$

Hence proved.

Find the value of $(\sin \theta + \sec \theta)^{2} + (\cos \theta + \text{cosec} \theta)^{2} - (1 + \sec \theta \text{cosec} \theta)^{2}$

$\left( \cos \theta +\csc \theta\right)^2 + \left(\sin \theta+\sec \theta\right)^2-(1+\sec \theta\csc \theta)^2$

$\implies \cos ^2 \theta+\csc ^2 \theta +2\cos \theta\csc \theta +\sin ^2 \theta +2\sin \theta\sec \theta +\sec ^2 \theta-(1+2\sec \theta\csc \theta+\csc^2 \theta)$

$\implies \left(\sin ^2 \theta+\cos ^2 \theta\right)+\csc ^2 \theta+\sec ^2 \theta-2 \left(\dfrac {\cos \theta}{\sin \theta} \right)+2 \left(\dfrac {\sin \theta}{\cos \theta}\right)-1-\sec^2 \theta\csc^2 \theta-2\sec \theta\csc \theta$

$\implies 1+\dfrac {1}{\sin ^2 \theta} +\dfrac {1}{\cos ^2 \theta}+2\left( \dfrac {\cos ^2 \theta+\sin ^2 \theta}{\sin \theta\cos \theta}\right)-1-2\sec \theta\csc \theta-\sec^2 \theta\csc^2 \theta$

$\implies \left( \dfrac {\cos ^2 \theta+\sin ^2 \theta}{\sin^2 \theta\cos^2 \theta}\right)+2\left( \dfrac {1}{\sin \theta\cos \theta}\right)-2\sec \theta\csc \theta-\sec^2 \theta\csc^2 \theta$

$\implies \left( \dfrac {1}{\sin^2 \theta\cos^2 \theta}\right)+2\sec \theta\csc \theta-2\sec \theta\csc \theta-\sec^2 \theta\csc^2 \theta$

$\implies \sec ^2 \theta\csc ^2 \theta-\sec^2 \theta\csc^2 \theta$

$\implies 0$

If $cosec^{6} \theta -cot^{6} = a cot^{4}\theta + b cot^{2}\theta + c$ then a + b + c =

We know that,

$a^3-b^3=(a-b)(a^2+ab+b^2)$

So,

$cosec^6\theta-\cot^6\theta=(cosec^2\theta-\cot^2\theta)(cosec^4\theta+cosec^2\theta.\cot^2\theta+\cot^4\theta)\\=(cosec^4\theta+cosec^2\theta.\cot^2\theta+\cot^4\theta)$

Since,

$cosec^2\theta-\cot^2\theta=1$

$cosec^4\theta+cosec^2\theta.\cot^2\theta+\cot^4\theta=cosec^2\theta(cosec^2\theta+\cot^2\theta)+\cot^4\theta=(1+\cot^2\theta)(1+2\cot^2\theta)+\cot^4\theta\\=1+3\cot^2\theta+3\cot^4\theta$

Hence,

$a+b+c=1+3+3=7$

If $\dfrac{\tan^3\alpha}{1+\tan^2\alpha}= \sec\alpha \sin\alpha- a \sin\alpha \cos\alpha$ . Find $a$

LHS

$\dfrac { { \tan }^{ 3 }\alpha }{ 1+{ \tan }^{ 2 }\alpha }$

$=\dfrac { \dfrac { { \sin }^{ 3 }\alpha }{ { \cos }^{ 3 }\alpha } }{ 1+\dfrac { { \sin }^{ 2 }\alpha }{ { \cos }^{ 2 }\alpha } } =\dfrac { \dfrac { { \sin }^{ 3 }\alpha }{ { \cos }^{ 3 }\alpha } }{ \dfrac { { \cos }^{ 2 }\alpha +{ \sin }^{ 2 }\alpha }{ { \cos }^{ 2 }\alpha } }$

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

$= \dfrac { { \sin }^{ 3 }\alpha { \cos }^{ 2 }\alpha }{ { \cos }^{ 3 }\alpha } =\dfrac { { \sin }^{ 3 }\alpha }{ \cos\alpha }$

$= \dfrac { \sin\alpha \left( { \sin }^{ 2 }\alpha \right) }{ \cos\alpha }$

$=\dfrac { \sin\alpha \left( 1-{ \cos }^{ 2 }\alpha \right) }{ \cos\alpha }$

$= \dfrac { \sin\alpha -\sin\alpha { \cos }^{ 2 }\alpha }{ \cos\alpha }$

$= \dfrac { \sin\alpha }{ \cos\alpha } -\sin\alpha \cos\alpha$

$= \sin\alpha \sec\alpha -\sin\alpha \cos\alpha$

$\Rightarrow \alpha =1$

If $x= r \cos\theta. \sin\phi, y= \sin\theta . \sin\phi, z= r \cos\phi$ Prove that $x^2+y^2+z^2=r^2$

$\textbf{Step 1: Substituting x y and z in the lhs part.}$

$\text{LHS}$

$={ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }$

$\text{putting value of x, y and z}$

$\text{LHS = }\quad { r }^{ 2 }{ \cos }^{ 2 }\theta { \sin }^{ 2 }\phi +{ r }^{ 2 }{ \sin }^{ 2 }\theta { \sin }^{ 2 }\phi +{ r }^{ 2 }{ \cos }^{ 2 }\phi$

$\quad \quad \textbf{[Given]}$

$= \quad { r }^{ 2 }{ \sin }^{ 2 }\phi \left( { \cos }^{ 2 }\theta +{ \sin }^{ 2 }\theta \right) +{ r }^{ 2 }{ \cos }^{ 2 }\phi$

$= \quad { r }^{ 2 }{ \sin }^{ 2 }\phi +{ r }^{ 2 }{ \cos }^{ 2 }\phi$

$= \quad { r }^{ 2 }\left( { \sin }^{ 2 }\phi +{ \cos }^{ 2 }\phi \right) \quad \quad \quad[ \because \quad { \sin }^{ 2 }x+{ \cos }^{ 2 }x=1]$

$= \quad { r }^{ 2 }$

$= \text{RHS}$

$\textbf{Hence proved.}$

Prove that $\sqrt { \dfrac { \sin { A } +1 }{ 1-\sin { A } } } +\sqrt { \dfrac { 1-\sin { A } }{ \sin { A } +1 } } =2\sec { A }$

$\sqrt{\dfrac{1+\sin A}{1-\sin A}}+\sqrt{\dfrac{1-\sin A}{1+\sin A}}$

$=\cfrac{\sqrt{1+\sin A}}{\sqrt{1-\sin A}}+\cfrac{\sqrt{1-\sin A}}{\sqrt{1+\sin A}}$

$=\dfrac{\left (\sqrt{1+\sin A} \right )^{2}+\left ( \sqrt{1-\sin A} \right )^{2}}{\sqrt{1-\sin ^{2}A}}$

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

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

$=2\sec A$

Solve: $\cos^{2}\theta +\cos^{2}\theta \cot^{2}\theta-\cot^{2}\theta$

We know that,

$\cot^2\theta=cosec^2\theta-1$

$\Rightarrow \cos^{2}\theta +\cos^{2}\theta \cot^{2}\theta-\cot^{2}\theta$

$=\cos^2\theta+\cos^2\theta(\csc^2\theta-1)-cot^2\theta$

$=\cos^2\theta+cot^2\theta-\cos^2\theta-\cot^2\theta$

$=0$

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

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

If $(\sec A+ \tan A -1)(\sec A-\tan A+1)= a\tan A$ .Find $a$

$\left( \sec{A} + \tan - 1 \right) \left( \sec{A} - \tan + 1 \right) = \left( \sec{A} + \left( \tan - 1 \right) \right) \left( \sec{A} - \left( \tan - 1 \right) \right)$

$\Rightarrow \; \sec^{2}{A} - {\left( \tan{A} - 1 \right)}^{2} \; \left\{ \because \left( a - b \right) \left( a + b \right) = {a}^{2} - {b}^{2} \right\}$

$\Rightarrow \; \sec^{2}{A} - \left( \tan^{2}{A} + 1 - 2 \tan{A} \right)$

$\Rightarrow \; \sec^{2}{A} - \left( \sec^{2}{A} - 2 \tan{A} \right) \; \left\{ \because 1 + \tan^{2}{A} = \sec^{2}{A} \right\}$

$\Rightarrow \;\sec^{2}{A} - \sec^{2}{A} + 2 \tan{A}$

$\Rightarrow \; 2 \tan{A}$

But given that-

$\left( \sec{A} + \tan - 1 \right) \left( \sec{A} - \tan + 1 \right) = a \tan{A}$

$\therefore \; a = 2$

Hence, the correct answer is 2.

If $\dfrac{\tan\theta}{(1+\tan^2\theta)^2}+ \dfrac{\cot\theta}{(1+ \cot^2\theta)^2}= m\sin \theta \cos\theta$ . Find $m$

$\cfrac{\tan{\theta}}{{\left( 1 + \tan^{2}{\theta} \right)}^{2}} + \cfrac{\tan{\theta}}{{\left( 1 + \tan^{2}{\theta} \right)}^{2}} = m \sin{\theta} \cos{\theta}$

$R.H.S.-$

$= m \sin{\theta} \cos{\theta}$

$L.H.S.-$

$\cfrac{\tan{\theta}}{{\left( 1 + \tan^{2}{\theta} \right)}^{2}} + \cfrac{\tan{\theta}}{{\left( 1 + \tan^{2}{\theta} \right)}^{2}} = \cfrac{\tan{\theta}}{{\left( \sec^{2}{\theta} \right)}^{2}} + \cfrac{\cot{\theta}}{{\left( \csc^{2}{\theta} \right)}^{2}} \; \left( \because \; 1 + \tan^{2}{\theta} = \sec^{2}{\theta} \; \& \; 1 + \cot^{2}{\theta} = \csc^{2}{\theta} \right)$

$\Rightarrow \; = \cfrac{\left( \cfrac{\sin{\theta}}{\cos{\theta}} \right)}{\left( \cfrac{1}{\cos^{4}{\theta}} \right)} + \cfrac{\left( \cfrac{\cos{\theta}}{\sin{\theta}} \right)}{\left( \cfrac{1}{\sin^{4}{\theta}} \right)}$

$\Rightarrow \; = \sin{\theta} \cos^{3}{\theta} + \cos{\theta} \sin^{3}{\theta}$

$\Rightarrow \; = \sin{\theta} \cos{\theta} \left( \sin^{2}{\theta} + \cos^{2}{\theta} \right)$

$\Rightarrow \; = \sin{\theta} \cos{\theta}$

On comparing the values of L.H.S. and R.H.S., we have

$m =1$

Hence, the correct answer is 1.

If $\sin\theta =\dfrac{3}{4}$ , show that $\sqrt{\dfrac{\text{cosec}^2\theta - \cot^2\theta}{\sec^2\theta-1}}=\dfrac{\sqrt m}{3}.$ .Find $m$

$\sin \theta=\cfrac {3}{4}$

$\sqrt {\cfrac {\csc^2\theta-\cot^2\theta}{\sec^2\theta-1}}=\cfrac {\sqrt{m}}{3}$ Find m.

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

$16=9+BC^2$

$\Rightarrow BC=\sqrt {7}$

(Refer to Image)

$\sin\theta=\cfrac {3}{4}=\cfrac {AB}{AC}$

$\csc\theta=\cfrac {AC}{AB}=\cfrac {4}{3}$

$\cot\theta=\cfrac {\sqrt 7}{3}=\cfrac {BC}{AB}$

$\sec\theta=\cfrac {AC}{BC}=\cfrac {4}{\sqrt7}$

$\Rightarrow\sqrt{\cfrac {\csc^2\theta-\cot^2\theta}{\sec^2\theta-1}}=$

$\sqrt {\cfrac {\left(\cfrac {4}{3}\right)^2-\left(\cfrac {\sqrt {7}}{3}\right)^2}{\left(\cfrac {4}{\sqrt {7}}\right)^2-1}}$

$=\sqrt {\cfrac {\cfrac {16}{9}-\cfrac {7}{9}}{\cfrac {16}{7}-1}}$

$=\sqrt {\cfrac {\cfrac{9}{9}}{\cfrac {9}{7}}}$

$=\sqrt {\cfrac {7}{9}}$

$=\cfrac{\sqrt{7}}{3}$

$\therefore m=7$

1006508_1044727_ans_7f915e9434f145029a3da3a7801c6493.PNG

$si{n^2}{30^ \circ } + co{s^2}60 + {\tan ^2}{45^ \circ } + se{c^2}{60^ \circ } - co{\sec ^2}{30^ \circ } = \frac{3}{2}$

$=sin^230^\circ + cos^2 60^\circ +tan^2 45^\circ +sec^2 60^\circ -cosec^2 30^\circ$

$= sin^230^\circ +sin^230^\circ +1 +cosec^230^\circ- cosec^2 30^\circ$

$= 2sin^230^\circ+1$

$= 2 *\dfrac{1}{4} +1$

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

If $\dfrac{\sin\theta - \cos\theta +1}{\sin\theta + \cos\theta -1}= \dfrac{m}{\sec\theta - \tan\theta},$ Find $m$

consider

$\cfrac { \sin { \theta } \cos { \theta } +1 }{ \sin { \theta } \cos { \theta } -1 } \\ \\$

multiply and divide by

$\cos\theta$

$\cfrac { \cfrac { \sin { \theta } \cos { \theta } +1 }{ \cos { \theta } } }{ \cfrac { \sin { \theta } \cos { \theta } -1 }{ \cos { \theta } } } =\cfrac { \cfrac { \sin { \theta } }{ \cos { \theta } } -1+\cfrac { 1 }{ \cos { \theta } } }{ \cfrac { \sin { \theta } }{ \cos { \theta } } +1-\cfrac { 1 }{ \cos { \theta } } } \\ =\cfrac { \tan { \theta } -1+\sec { \theta } }{ \tan { \theta } +1-\sec { \theta } }$

put

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

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

multiply and divide by

$\sec\theta -\tan\theta$

$=\tan { \theta } +\sec { \theta } \times \cfrac { \tan { \theta } -\sec { \theta } }{ \tan { \theta } -\sec { \theta } } \\$

$=\cfrac { 1 }{ \tan { \theta } -\sec { \theta } }$ [since

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

$=m=1$

If $x= \text{cosec}A + \cos A$ and $y= \text{cosec}A- \cos A$ then prove that $\left(\dfrac{2}{x+y}\right) ^2+ \left(\dfrac{x-y}{2}\right)^2-1=0$ .

Given:-

$\Rightarrow$

$x = cosec{A} + \cos{A}$

$\Rightarrow$

$y = cosec{A} - \cos{A}$

To prove:-

$\Rightarrow$

${\left( \cfrac{2}{x + y} \right)}^{2} + {\left( \cfrac{x - y}{2} \right)}^{2} - 1 = 0$

$\therefore \; x + y = cosec{A} + \cos{A} + cosec{A} - \cos{A} = 2 cosec{A}$

$x - y = cosec{A} + \cos{A} - cosec{A} + \cos{A} = 2 \cos{A}$

$\therefore \; {\left( \cfrac{2}{x + y} \right)}^{2} + {\left( \cfrac{x - y}{2} \right)}^{2} - 1 = {\left( \cfrac{2}{2 cosec{A}} \right)}^{2} + {\left( \cfrac{2 \cos{A}}{2} \right)}^{2} - 1$

$\Rightarrow \; {\left( \sin{A} \right)}^{2} + {\left( \cos{A} \right)}^{2} - 1$

$\Rightarrow \; \sin^{2}{A} + \cos^{2}{A} - 1$

$\Rightarrow \; 1 - 1$

$\Rightarrow \; 0$

Hence proved.

If ${\cos}^{4}{\theta}+{\cos}^{2}{\theta}=1$ then show that
i) ${\sec}^{4}{\theta}-{\sec}^{2}{\theta}=1$
ii) ${\cot}^{4}{\theta}-{\cot}^{2}{\theta}=1$

Given:

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

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

${\cos ^4}\theta = {\sin ^2}\theta \,\,\,\,\,\,\,\,\,\left( 1 \right)$

Now,

$\left( 1 \right)\,\,\,\,\,{\sec ^4}\theta - {\sec ^2}\theta$

$= \cfrac{1}{{{{\cos }^4}\theta }} - \cfrac{1}{{{{\cos }^2}\theta }}$

$= \cfrac{{ 1- {{\cos }^2}\theta }}{{{{\cos }^4}\theta }}$

$= \cfrac{{{{\sin }^2}\theta }}{{{{\cos }^4}\theta }}$

$= \cfrac{{{{\sin }^2}\theta }}{{{{\sin }^2}\theta }}\,\,\,\,\,\,\,\,\left( {from\left( 1 \right)} \right)$

$=1$

$\left( 2 \right)\,\,\,\,{\cot ^4}\theta - {\cot ^2}\theta$

$= \cfrac{{{{\cos }^4}\theta }}{{{{\sin }^4}\theta }} - \cfrac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}$

$= \cfrac{{{{\cos }^4}\theta - {{\sin }^2}\theta {{\cos }^2}\theta }}{{{{\sin }^4}\theta }}$

$= \cfrac{{{{\sin }^2}\theta - {{\sin }^2}\theta {{\cos }^2}\theta }}{{{{\sin }^4}\theta }}\,\,\,\left[ {from\,\left( 1 \right)} \right]$

$= \cfrac{{{{\sin }^2}\theta \left( {1 - {{\cos }^2}\theta } \right)}}{{{{\sin }^4}\theta }}$

$= \cfrac{{{{\sin }^2}\theta \times {{\sin }^2}\theta }}{{{{\sin }^4}\theta }}$

$= \cfrac{{{{\sin }^4}\theta }}{{{{\sin }^4}\theta }}$

$=1$

If $A+ B =90^{\circ}$ then prove $\displaystyle \sqrt{\frac{ \tan A\tan B + \tan A\cot B}{\sin A \sec B} -\frac{\sin^{2}B}{\cos^{2}A}} = \tan A$

Given that,

$A + B = 90^{\circ}$

$\implies B = 90^{\circ} - A$

$\tan B = \tan(90^{\circ}-A) = \cot A$

$\sec B = \sec(90^{\circ}-A)= \csc A$

$\sin B = \sin(90^{\circ}-A)= \cos A$

$\therefore \sqrt{\dfrac{\tan A\tan B + \tan B\tan A}{\sec A\sec B} - \dfrac{\sin^2 B}{\cos^2 A}}$

$= \sqrt{\dfrac{\tan A \cot A + \cot A\tan A}{\sin A\csc A} - \dfrac{\cos^2 A}{\cos^2 A}}$

$=\sqrt{\dfrac{1+1}{1}-1}$

$= \sqrt{2-1} = \sqrt{1} = 1$

Solve $7cos^{2}\theta + 3 sin^{2}\theta =4$ .

$7\cos^2{\theta} +3\sin^2{\theta} = 4$

$4\cos^2{\theta} + 3\cos^2{\theta} + 3\sin^2{\theta} = 4$

$4\cos^2{\theta} +3(1) = 4$

$4\cos^2{\theta} = 4-3=1$

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

$\cos \theta = \dfrac{1}{2} = \cos 60^{\circ}$

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

Formula:

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

Evaluate
$\dfrac{{\sin {{90}^ \circ }}}{{\cos {{45}^ \circ }}} + \dfrac{1}{{\cos ec{{30}^ \circ }}}$

We know that

$\sin{{90}^{\circ}}=1\,\,,\cos{{45}^{\circ}}=\dfrac{1}{\sqrt{2}}\,\,,\csc{{30}^{\circ}}=2$

$\dfrac{\sin{{90}^{\circ}}}{\cos{{45}^{\circ}}}+\dfrac{1}{\csc{{30}^{\circ}}}$

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

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

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

Simplifying the LHS of $\cfrac{{1 + \cos A + \sin A}}{{1 + \cos A - \sin A}} = \cfrac{{1 + \sin A}}{{\cos A}}$ .

$\cfrac{{1 + \cos A + \sin A}}{{1 + \cos A - \sin A}} = \cfrac{{1 + \cos A + \sin A}}{{1 + \cos A - \sin A}} \times \cfrac{{1 + \cos A + \sin A}}{{1 + \cos A + \sin A}}$

$= \cfrac{{{{\left( {1 + \cos A + \sin A} \right)}^2}}}{{{{\left( {1 + \cos A} \right)}^2} - {{\sin }^2}A}}$

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

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

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

$= \cfrac{{2\left( {1 + \cos A} \right) + 2\sin A\left( {1 + \cos A} \right)}}{{2\cos A + 2{{\cos }^2}A}}$

$= \cfrac{{2\left( {1 + \cos A} \right)\left( {1 + \sin A} \right)}}{{2\cos A\left( {1 + \cos A} \right)}}$

$= \cfrac{{1 + \sin A}}{{\cos A}}$

$= {\rm{RHS}}$

If $\tan\theta =\dfrac{1}{\sqrt{7}}$ , show that $\dfrac{cosec^2\theta -\sec^2\theta}{cosec^2\theta +\sec^2\theta}=\dfrac{3}{4}$ .

$\tan\theta=\cfrac{1}{\sqrt7},\\ \cfrac{\sin\theta}{\cos\theta}=\cfrac{1}{\sqrt 7},\\ \cfrac{\sin^2\theta}{\cos^2\theta}=\cfrac{1}{7}, \cos^2\theta+\sin^2\theta=1$

$\cos^2=7\sin^2\theta, 7\sin^2\theta+\sin^2\theta=1\Rightarrow=\cfrac{1}{8}\\ \therefore \cos^2\theta=1-\sin^2\theta=1-\cfrac{1}{8}=\cfrac{7}{8}\\ \therefore cosec^2\theta=\cfrac{1}{\sin^2\theta}=8\\ \therefore\sec^2=\cfrac{1}{\cos^2\theta}=\cfrac{8}{7}\\ \therefore\cfrac{cosec^2\theta-\sec^2\theta}{cosec^2\theta+\sec^2\theta}=\cfrac{8-{\cfrac{8}{7}}}{8+{\cfrac{8}{7}}}\\=\cfrac{3}{4}$

Henceproved

$2x\tan ^{ 2 }{ { 60 }^{ o }+3x\sin ^{ 2 }{ { 30 }^{ o } } } =\dfrac { 27\cos ^{ 2 }{ { 45 }^{ o } } }{ 4\sin ^{ 2 }{ { 60 }^{ o } } }$

Given that,

$2x{{(tan{{60}^{\circ }})}^{2}}+3x{{(sin{{30}^{\circ }})}^{2}}=\dfrac{27{{(cos{{45}^{\circ }})}^{2}}}{4{{(sin{{60}^{\circ }})}^{2}}}$

On substituting the values, we get

$2x{{(\sqrt{3})}^{2}}+3x{{(\dfrac{1}{2})}^{2}}=\dfrac{27{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}}{4{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}}$

$6x+\dfrac{3x}{4}=\dfrac{9}{2}$

$2(24x+3x)=36$

$54x=36$

$x=\dfrac{2}{3}$

Hence, this is the answer.

Prove: $1- \dfrac{cos^2A}{1+SinA}= SinA.$

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

Prove: $\cfrac{{\cot \theta - \cos \theta }}{{\cot \theta + \cos \theta }} = \cfrac{{{\mathop{\rm cosec}\nolimits} \theta - 1}}{{{\mathop{\rm cosec}\nolimits} \theta + 1}}$

L.H.S

$\Rightarrow \cfrac{{\cot \theta - \cos \theta }}{{\cot \theta + \cos \theta }}$

$\Rightarrow \cfrac{{\dfrac{\cos \theta}{\sin \theta} - \cos \theta }}{{\dfrac{\cos \theta}{\sin \theta} + \cos \theta }}$

$\Rightarrow \cfrac{{\dfrac{1}{\sin \theta} - 1}}{{\dfrac{1}{\sin \theta} + 1}}$

$\Rightarrow \cfrac{{\csc \theta- 1}}{{\csc \theta + 1}}$

R.H.S

Hence, proved.

If $\cot x =2$ , find the value of $\dfrac{(2+2\sin x)(1-\sin x)}{(1+\cos x)(2-2\cos x)}$

$\cot x=2$

$=\cfrac { 2(1+\sin x)(1-\sin x) }{ 2(1-\cos x)(1+\cos x) }$

$=\cfrac { 1-{ \sin }^{ 2 }x }{ 1-{ \cos }^{ 2 }x }$

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

$={ \cot }^{ 2 }x$

$=(2)^2$

$=4$

$\dfrac{2}{3}\text{cosec}^258^0 -\dfrac{2}{3}\cot 58^0 \tan32^0-\dfrac{5}{3}\tan 13^0\tan37^0\tan45^0\tan53^0\tan 77^0=-m$ .Find $m$

$\cfrac{2}{3}cosec^2 58°-\cfrac{2}{3} \cot 58°.\tan 32°-\cfrac{5}{3}\tan13°.\tan37°.\tan45°.\tan53°.\tan77°$

$\implies \cfrac{2}{3} cosec^2 58°-\cfrac{2}{3}\cot^2 58°-\cfrac{5}{3}=\cfrac{2}{3}(cosec^2 58°-\tan^2 58°)-\cfrac{5}{3}$

$\implies \cfrac{2}{3}-\cfrac{5}{3}=\cfrac{2-5}{3}=\cfrac{-3}{3}=-1$

${\sec ^4}x - {\sec ^2}x = {\tan ^4}x + {\tan ^2}x$

$L.H.S. = \sec^2x(sec^2x -1)$

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

$R.H.S. =\ tan^2x(\tan^2x+1)$

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

Hence,

$L.H.S. = R.H.S.$ for all values of x

Prove: $\left( {1 + \,{{\tan }^2}\,\theta } \right)\,{\sin ^2}\,\theta \, = \,{\tan ^2}\,\theta$

We know that

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

L.H.S

$={\sec}^{2}{\theta}{\sin}^{2}{\theta}$

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

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

$={\tan}^{2}{\theta}=$ R.H.S

Hence proved.

Prove: $\sec ^{ 4 }{ \theta } -\tan ^{ 4 }{ \theta } =2\sec ^{ 2 }{ \theta } -1$

$LHS= (\sec^{4}\theta -\tan^{4}\theta )$

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

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

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

$=RHS$

Prove that $\dfrac{\sin \theta-\cos \theta+1}{\sin \theta+ \cos \theta-1}=\dfrac{1}{\sec \theta- \tan \theta }$ [use the identity $\sec^{2} \theta=1+\tan^{2} \theta$ .]

L.H.S

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

$=\dfrac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +1}=\dfrac{\left( \tan \theta +\sec \theta \right)-\left( {{\sec }^{2}}\theta -{{\tan }^{2}}\theta \right)}{\tan \theta -\sec \theta +1}$

$=\dfrac{\left( \tan \theta +\sec \theta \right)-\left( \sec \theta -\tan \theta \right)\left( \sec \theta +\tan \theta \right)}{\tan \theta -\sec \theta +1}$

$=\dfrac{\left( \tan \theta +\sec \theta \right)\left( 1-\sec \theta +\tan \theta \right)}{\left( 1-\sec \theta +\tan \theta \right)}$

$=\sec \theta +\tan \theta$

Multiplying and Dividing by $\sec \theta -\tan \theta$ we get,

$=\dfrac{{{\sec }^{2}}\theta -{{\tan }^{2}}\theta }{\sec \theta -\tan \theta }$

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

Prove that $\dfrac{2}{1-\sin A}\times \dfrac{1}{1+\sin A}=2\sec^2 A$

L.H.S

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

We know that

${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$

Therefore,

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

$\Rightarrow \dfrac{2}{{{\cos }^{2}}A}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \because 1-{{\sin }^{2}}x={{\cos }^{2}}x \right]$

$\Rightarrow 2{{\sec }^{2}}A$

R.H.S

Hence, proved.

Evaluate $\dfrac {2\cos^{2}90^{o}+4\cos^{2}45^{o}+\tan^{2}60^{o}+3csc^{2}60^{o}+1}{3csc 60^{o}-\dfrac {7}{2}\sec^{2}45^{o}+2\csc 30^{o}-1}$

Consider the following question.

$\dfrac{2{{\cos }^{2}}{{90}^{0}}+4{{\cos }^{2}}{{45}^{0}}+{{\tan }^{2}}{{60}^{0}}+3{{\csc }^{2}}{{60}^{0}}+1}{3\csc {{60}^{0}}-\dfrac{7}{2}{{\sec }^{2}}45+2\csc {{30}^{0}}-1}$

$=\dfrac{2*0+4{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}+{{\left( \sqrt{3} \right)}^{2}}+3{{\left( \dfrac{2}{\sqrt{3}} \right)}^{2}}+1}{3{{\left( \dfrac{2}{\sqrt{3}} \right)}^{2}}-\dfrac{7}{2}{{\left( \sqrt{2} \right)}^{2}}-1}$

$=-2$

Hence, this is the required answer.

If $\sec{\theta}=x+\cfrac{1}{4x}$ then prove $\sec{\theta}+\tan{\theta}=2x$

Consider

$sec\;\theta=x+\dfrac{1}{4x}$

We know that

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

$\Rightarrow tan^2\theta=\left(x+\dfrac{1}{4x}\right)^2-1$

$=x^2+\dfrac{1}{16x^2}+\dfrac{1}{2}-1$

$=x^2+\dfrac{1}{16x^2}-\dfrac{1}{2}$

$=x^2+\dfrac{1}{16x^2}-2\cdot x\cdot \dfrac{1}{4x}$

$=\left(x-\dfrac{1}{4x}\right)^2$

$\Rightarrow tan\:\theta=x-\dfrac{1}{4x}$

$\therefore sec\:\theta+tan\:\theta=x+\dfrac{1}{4x}+x-\dfrac{1}{4x}$

$\Rightarrow sec\:\theta+tan\:\theta=2x$

Evaluate:
${ \sin }^{ 2 }{ 5 }^{ o }+{ \sin }^{ 2 }{ 10 }^{ o }+..........{ \sin }^{ 2 }{ 85 }^{ o }$ ?

$=sin^{ 2 }5+sin^{ 2 }10+……+sin^{ 2 }75+sin^{ 2 }80+sin^{ 2 }85+$

$=(sin^{ 2 }5+sin^{ 2 }85)+(sin^{ 2 }10+sin^{ 2 }80)+……..+sin^{ 2 }44+...$

$=(sin^{ 2 }5+cos^{ 2 }5)+(sin^{ 2 }10+cos^{ 2 }10)+……..$

$=(1+1+1+1 =...40 times)+\dfrac { 1 }{ 2 }$

$=40+\dfrac { 1 }{ 2 }\\$

$=40.5$

Prove: ${ \left( \tan { \theta } +\sec { \theta } \right) }^{ 2 }+{ \left( \tan { \theta } -\sec { \theta } \right) }^{ 2 }=\cfrac { 2\left( 1+\sin ^{ 2 }{ \theta } \right) }{ \cos ^{ 2 }{ \theta } }$

$LHS= \tan^{2}\theta +\sec^{2}\theta +2 \tan\theta \sec\theta + \tan^{2}\theta +\sec^{2}\theta -2 \tan\theta \sec\theta$

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

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

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

$\tan^{2}{\phi}-\sin^{2}{\phi}=\tan^{2}{\phi}.\sin^{2}{\phi}$

Solving LHS,

$tan^2\phi-sin^2\phi$

$\dfrac{sin^2\phi}{cos^2\phi}-sin^2\phi$

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

$sin^2\phi(sec^2\phi-1)$

$sin^2\phi .tan^2\phi$ = RHS

Hence proved,

Solve :
$2\tan^{2}{45^{0}}+\cos^{2}{45^{0}}-\sin^{2}{45^{0}}$

1348541_1134768_ans_bca8164b25e2405ca66d44e2d1c9a79a.jpg

Solve
${\sec ^6}x - {\tan ^6}x - 3{\sec ^2}x.{\tan ^2}$

We know that

$a^3-b^3=(a-b)(a^2+ab+b^2)$

Now,

$\sec^6x-\tan^6x=(\sec^2x-\tan^2x)(\sec^4x+\sec^2x\tan^2x+\tan^4x)=1.(\sec^4x+\sec^2x\tan^2x+\tan^4x)\\=(\sec^2x-\tan^2x)^2+2\sec^2x\tan^2x+\tan^2x\sec^2x\\=1+3\sec^2x\tan^2x$

Hence,

$\sec^6x-\tan^6x-3\sec^2x\tan^2x\\=1+3\sec^2x\tan^2x-3\sec^2x\tan^2x=1$

To prove that
$\left(\dfrac {1}{\sec^{2}x-\cos^{2}x}+\dfrac {1}{\csc^{2}x-\sin^{2}x}\right)\sin^{2}x.\cos^{2}x$
$=\dfrac {1-\sin^{2}x \cos^{2}x}{2+\sin x\cos^{2}x}$

$\left[ \dfrac { 1 }{ \left( \sec ^{ 2 }{ x } -\cos ^{ 2 }{ x } \right) } +\dfrac { 1 }{ \left( \csc ^{ 2 }{ x } -\sin ^{ 2 }{ x } \right) } \right] \sin ^{ 2 }{ x } \cos ^{ 2 }{ x }$

$=\left[ \dfrac { \cos ^{ 2 }{ x } }{ 1-\cos ^{ 4 }{ x } } +\dfrac { \sin ^{ 2 }{ x } }{ 1-\sin ^{ 4 }{ x } } \right] \sin ^{ 2 }{ x } \cos ^{ 2 }{ x }$

$=\left[ \dfrac { \cos ^{ 2 }{ x } -\cos ^{ 2 }{ x } \sin ^{ 2 }{ x } -\sin ^{ 2 }{ x } \cos ^{ 4 }{ x } }{ 1+\sin ^{ 4 }{ x } \cos ^{ 4 }{ x } -\cos ^{ 4 }{ x } -\sin ^{ 4 }{ x } \sin ^{ 2 }{ x } \cos ^{ 2 }{ x } } \right]$

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

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

Find the values of the trigonometric ratio
$\tan \theta=11$

According to question,

$tanA=\dfrac{11}{1}=\dfrac{P}{B}$

Let hypotenuse ,

$H=3k$

So, perpendicular,

$P=2k$

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

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

$=(11k)^2+(1k)^2$

$=122k^2$

$H =\sqrt{122}k$

$cosA=\dfrac{B}{H}$

$= \dfrac{1}{\sqrt{122}}$

$cotA=\dfrac{B}{P}$

$= \dfrac{1}{11}$

$cosecA=\dfrac{H}{P}$

$= \dfrac{\sqrt{122}}{11}$

$secA=\dfrac{H}{B}$

$= \dfrac{\sqrt{122}}{1}$

$sinA=\dfrac{P}{H}$

$= \dfrac{11}{\sqrt{122}}$

Show that $(\text{cosec}{\theta}-\cot{\theta})^{2}=\dfrac{1-\cos{\theta}}{1+\cos{\theta}}$

L.H.S

$=(\text{cosec} \theta -\cot \theta )^2$

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

$[\because \text{cosec} \theta=\dfrac{1}{\sin \theta}$ and

$\cot \theta=\dfrac{\cos\theta}{\sin \theta}]$

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

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

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

$[\because \sin^2 \theta=1-\cos^2 \theta]$

$= \ \dfrac {(1-\cos \theta)\times (1-\cos \theta)}{(1+\cos \theta)\times (1-\cos \theta)}$

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

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

$=$ R.H.S

$\dfrac{\cos{A}-1}{2-\sec^{2}{A}}=\dfrac{\cot{A}}{1+\tan{A}}$
Find A.

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

$\dfrac{\cos A-1}{(1-\tan A)(1+\tan A)}=\dfrac{\cot A}{(1+\tan A)}$

$\cos A-1=\cot A-1$

$\cos A=\dfrac{\cos A}{\sin A}$

$\sin A=1$

$\boxed{A=90^o}$

If $x=a\sin\theta$ and $y=b\tan\theta$ , then prove that $\dfrac{a^{2}}{x^{2}}-\dfrac{b^{2}}{y^{2}}=1$ .

$x=a\sin \theta$

$\therefore\dfrac{a}{x}=\csc\theta$

$\therefore y=b\tan\theta$

$\therefore \dfrac{b}{y}=\cot\theta$

$\therefore \dfrac{a^{2}}{x^{2}}-\dfrac{b^{2}}{y^{2}}=\csc^{2}\theta-\cot^{2}\theta$

$\therefore \dfrac{a^{2}}{x^{2}}-\dfrac{b^{2}}{y^{2}}=1$

Express $\sec{\theta}$ in terms of $\cot{\theta}$

We know that

$sec\;\theta=\dfrac{1}{cos\;\theta}$

We can convert it into

$=\dfrac{sin\;\theta}{cos\;\theta}.\dfrac{1}{sin\;\theta}$

$=\dfrac{1}{cot\;\theta}cosec\;\theta$

So,

$sec\;\theta=\dfrac{1}{cot\;\theta}\left (1+cot\;\theta\right )^{\frac{1}{2}}$

If $x=r\sin A\cos C, y=r\sin A\sin C$ and $z=r\cos A$ , prove that $r^{2}=x^{2}+y^{2}+z^{2}$ .

$x=r\sin A\cos C$

$y=r\sin A\ \ \ \ \sin C$

$\therefore x^{2}+y^{2}=r^{2}\sin^{2}A\cos^{2}C+r^{2}\sin^{2}A\sin^{2}C$

$\therefore x^{2}+y^{2}=r^{2}\sin^{2}A(\cos^{2}C+\sin^{2}C)$

$\therefore x^{2}+y^{2}=r^{2}\sin^{2}A$

$\therefore x^{2}+y^{2}+z^{2}=r^{2}\sin^{2}A+r^{2}\cos^{2}A$

$\therefore x^{2}+y^{2}+z^{2}=r^{2}(\sin^{2}+\cos^{2}A)$

$\therefore x^{2}+y^{2}+z^{2}=r^{2}$

Hence proved

If $\dfrac{\cos{A}}{\cos{B}}=m$ and $\dfrac{\cos{A}}{\sin{B}}=n.$ Prove:- $\left(m^{2}+n^{2}\right)\cos^{2}B=n$

${ m }^{ 2 }=\cfrac { \cos ^{ 2 }{ A } }{ \cos ^{ 2 }{ B } } ;{ n }^{ 2 }=\cfrac { \cos ^{ 2 }{ A } }{ \sin ^{ 2 }{ B } } ...(1)$

$\quad \therefore { m }^{ 2 }+{ n }^{ 2 }=\cfrac { \cos ^{ 2 }{ A } }{ \sin ^{ 2 }{ B } \cos ^{ 2 }{ B } }$

$\left( { m }^{ 2 }+{ n }^{ 2 } \right) \cos ^{ 2 }{ B } =\cfrac { \cos ^{ 2 }{ A } }{ \sin ^{ 2 }{ B } } ...(2)$

from (1) and (2)

$\left( { m }^{ 2 }+{ n }^{ 2 } \right) \cos ^{ 2 }{ B } ={ n }^{ 2 }$

Show that $\dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = 1 + \sec A cosecA$

L.H.S

$\Rightarrow \dfrac{\tan A}{1-\cot A}+\dfrac{\cot A}{1-\tan A}$

$\Rightarrow \dfrac{\dfrac{\sin A}{\cos A}}{\dfrac{\sin A-\cos A}{\sin A}}+\dfrac{\dfrac{\cos A}{\sin A}}{\dfrac{\cos A-\sin A}{\cos A}}$

$\Rightarrow \dfrac{{{\sin }^{2}}A}{\cos A\left( \sin A-\cos A \right)}+\dfrac{{{\cos }^{2}}A}{\sin A\left( \cos A-\sin A \right)}$

$\Rightarrow \dfrac{{{\sin }^{2}}A}{\cos A\left( \sin A-\cos A \right)}-\dfrac{{{\cos }^{2}}A}{\sin A\left( \sin A-\cos A \right)}$

$\Rightarrow \dfrac{{{\sin }^{3}}A-{{\cos }^{3}}A}{\cos A\sin A\left( \sin A-\cos A \right)}$

$\Rightarrow \dfrac{\left( \sin A-\cos A \right)\left( {{\sin }^{2}}A+{{\cos }^{2}}A+\cos A\sin A \right)}{\cos A\sin A\left( \sin A-\cos A \right)}$

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

$\Rightarrow 1+\sec A\csc A$

Hence, proved.

Prove that $\dfrac { 1 + \cos A } { \sin A } + \dfrac { \sin A } { 1 + \cos A } = 2 \csc A$

LHS =

$\frac{1+cosA}{sinA}+\frac{sinA}{1+cosA}$

$= \frac{(1+cosA)^{2}+sin^{2}A}{sinA(1+cosA)}$

$= \frac{1+2cosA+cos^{2}A+sin^{2}A}{sinA(1+cosA)}$

$= \frac{1+2cosA+1}{sinA(1+cosA)}[sin^{2}A+cos^{2}A = 1 ]$

$= \frac{2+2cosA}{sinA(1+cosA)}$

$= \frac{2[1+cosA]}{sinA(1+cosA)}$

$= \frac{2}{sinA}$

$= 2 cosec A = RHS$

LHS = RHS

Hence proved.

1090190_1180528_ans_9e06c8cdd8be4668897ab8e11c03f2ba.png

Evaluate:
$\tan ^{ 2 }{ { 30 }^{ o } } \sin { { 30 }^{ o } } +\cos { { 60 }^{ o } } \sin ^{ 2 }{ { 90 }^{ o } } \tan ^{ 2 }{ { 60 }^{ o } } -2\tan { { 45 }^{ o } } \cos ^{ 2 }{ { 0 }^{ o } } \sin { { 90 }^{ o } }$

$\tan^230\sin30+\cos60\sin^290\tan^260-2\tan45\cos^20\sin90$

$=\left(\dfrac{1}{\sqrt3}\right)^2\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)(1)^2(\sqrt3)^2-2(1)(1)^2(1)$

$=\dfrac{1}{6}+\dfrac{3}{2}-2$

$=\dfrac{1+9-12}{6}$

$=\dfrac{-2}{6}$

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

Given $sec\: \theta = \dfrac {13}{12}$ , calculate all other trigonometric ratios

$\sec{\theta}=\dfrac{13}{12}\Rightarrow \cos{\theta}=\dfrac{1}{\sec{\theta}}=\dfrac{12}{13}$

$\sin{\theta}=\sqrt{1-{\cos}^{2}{\theta}}=\sqrt{1-{\left(\dfrac{12}{13}\right)}^{2}}=\sqrt{1-\dfrac{144}{169}}=\sqrt{\dfrac{25}{169}}=\pm\dfrac{5}{13}$

Note

$\cos{\theta}$ is positive in first and fourth quadrant and

$\sin{\theta}$ is positive in first quadrant and negative in third quadrant.

$\csc{\theta}=\dfrac{1}{\sin{\theta}}=\dfrac{1}{\dfrac{\pm 5}{13}}=\pm\dfrac{13}{5}$

$\tan{\theta}=\dfrac{\sin{\theta}}{\cos{\theta}}=\dfrac{\pm\dfrac{13}{5}}{\dfrac{12}{13}}=\dfrac{\pm 13}{12}$

$\cot{\theta}=\dfrac{1}{\tan{\theta}}=\pm\dfrac{12}{13}$

In fig. $\triangle PQR$ right angled at $Q,PQ=6cm$ and $PR=12cm$ . Determine $\angle QPR$ and $\angle PRQ$ .

Given

$PQ=6\ cm$

$PR=12\ cm$

$\sin R=\dfrac{PQ}{PR}$

$=\dfrac{6}{12}=\dfrac{1}{2}$

$\sin R=\sin 30^o$

$R=30^o$

$\angle PRQ=30^o$

$\angle P+\angle Q+\angle R=180^o$

$\angle P+90^o+30^o=180^o$

$\angle P=180^o-120^o$

$\angle P=60^o$

$\angle QPR=60^o$ &

$\angle PRQ=30^o$

1347948_1231926_ans_a229fd6661fe4945a12d114a69460d6c.png

If $\tan \theta + \sec \theta =1.5$ , find $\sin \theta$ .

1643514_1221981_ans_e389c6f15e524553851e9508a8b47657.png

Evaluate $\dfrac { \sin 18 ^ { \circ } } { \cos 72 ^ { \circ } }$ .

$\dfrac { \sin 18 ^ { \circ } } { \cos 72 ^ { \circ } } = \dfrac { sin 18^0}{ cos ( 90 - 18)}$

$= \dfrac { sin 18^0}{ sin 18^0} = 1$

Prove that: $\cfrac { \sin { \theta } }{ \left( 1-\cot { \theta } \right) } +\cfrac { \cos { \theta } }{ \left( 1-\tan { \theta } \right) } =\cos { \theta } +\sin { \theta }$

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

$=\dfrac{\sin\theta}{1-(1/\tan\theta)}+\dfrac{\cos\theta}{1-\tan\theta}$

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

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

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

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

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

$=\dfrac{(\sin\theta+\cos\theta)(\sin\theta-\cos\theta)}{\sin\theta-\cos\theta)}$

$=\sin\theta+\cos\theta$

Evaluate : $\dfrac{ cos 22^0}{sin 68^0}$

1521158_1232638_ans_8aecf1fa393642da9b23607f03e597cd.jpg

If $4tan\theta =3$ , evaluate $\left( \dfrac { 4sin\theta -cos\theta +1 }{ 4sin\theta +cos\theta -1 } \right)$

1306828_1407875_ans_8a796648f56142318d6141795189bfc5.JPG

If $\cot\theta=3x-\dfrac{1}{12x}$ then show that $\csc\theta+\cot\theta=6x$ or $-\dfrac{1}{6x}$ .

$\cot{\theta} = 3x - \cfrac{1}{12x}$

Squaring both sides, we have

$\cot^{2}{\theta} = {\left( 3x - \cfrac{1}{12x} \right)}^{2}$

$\csc^{2}{\theta} - 1 = 9{x}^{2} + \cfrac{1}{144 {x}^{2}} - \cfrac{1}{2}$

$\Rightarrow \csc^{2}{\theta} = 9{x}^{2} + \cfrac{1}{144 {x}^{2}} - \cfrac{1}{2} + 1$

$\Rightarrow \csc^{2}{\theta} = 9{x}^{2} + \cfrac{1}{144 {x}^{2}} + \cfrac{1}{2}$

$\Rightarrow \csc^{2}{\theta} = {\left( 3x + \cfrac{1}{12x} \right)}^{2}$

$\Rightarrow \csc{\theta} = \pm \left( 3x + \cfrac{1}{12x} \right)$

Now,

Case I:-

$\csc{\theta} = 3x + \cfrac{1}{12x}$

Therefore,

$\cot{\theta} + \csc{\theta} = \left( 3x - \cfrac{1}{12x} \right) + \left( 3x + \cfrac{1}{12x} \right) = 6x$

Case II:-

$\csc{\theta} = - \left( 3x + \cfrac{1}{12x} \right)$

Therefore,

$\cot{\theta} + \csc{\theta} = \left( 3x - \cfrac{1}{12x} \right) + \left( -3x - \cfrac{1}{12x} \right) = - \cfrac{1}{6x}$

Hence proved.

Evaluate : $\dfrac { 4 }{ { cot }^{ 2 }{ 30 }^{ 0 } } +\dfrac { 1 }{ { sin }^{ 2 }{ 30 }^{} } -2{ cos }^{ 2 }{ 45 }^{ 0 }-{ sin }^{ 2 }{ 0 }^{ 0 }$

$\dfrac{4}{{\cot}^{2}{{30}^{\circ}}}+\dfrac{1}{{\sin}^{2}{{30}^{\circ}}}-2{\cos}^{2}{{45}^{\circ}}-{\sin}^{2}{{0}^{\circ}}$

$=\dfrac{4}{{\left(\sqrt{3}\right)}^{2}}+\dfrac{1}{{\left(\dfrac{1}{2}\right)}^{2}}-2{\left(\dfrac{1}{\sqrt{2}}\right)}^{2}-0$

$=\dfrac{4}{3}+4-2\times \dfrac{1}{2}$

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

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

$=\dfrac{4+9}{3}=\dfrac{13}{3}$

If $\cot \theta=\dfrac {15}{8}$ , find $\cos \theta$ and $\csc \theta$ .

1643541_1414154_ans_aaeea483a80a4767bcb8eed4003c4c00.PNG

Find the value of $6\tan^{ 2 }\theta -6\sec^{ 2 }\theta$ .

$\textbf{Step 1: Apply relevant identities of trigonometry.}$

$\text{To find the value of }$

$(6\tan^{ 2 }\theta -6\sec^{ 2 }\theta)$

$\text{Now, solving the given expression we get,}$

$\Rightarrow 6\tan^{ 2 }\theta -6\sec^{ 2 }\theta$

$=6(\tan^{ 2 }\theta -\sec^{ 2 }\theta)$

$=6\times(- 1)$

$\left[\boldsymbol{\because \sec^{ 2 }\theta-\tan^{ 2 }\theta =1}\right]$

$=-6$

$\textbf{Hence, the required value is }$

$-6 .$

Evaluate :
$\left( 3-\cot{ 30 }^{ \circ } \right) -{ \tan }^{ 3 }{ 60 }^{ \circ }+2\tan{ 60 }^{ \circ }$

We know that

$\cot{{30}^{\circ}}=\sqrt{3},\,\,\tan{{60}^{\circ}}=\sqrt{3}$

$3-\cot{{30}^{\circ}}-{\tan}^{3}{{60}^{\circ}}+2\tan{{60}^{\circ}}$

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

$=3-\sqrt{3}-3\sqrt{3}+2\sqrt{3}$

$=3-2\sqrt{3}$

Prove that :
$(\sin\theta+\cos\theta)(\tan\theta+\cot\theta)=\sec\theta+\csc\theta$ .

$\textbf{Step 1: Apply suitable formula to prove the desired identity.}$

$\text{On solving LHS, we get,}$

$\text{L.H.S }=\left( \sin{\theta} + \cos{\theta} \right) \left( \tan{\theta} + \cot{\theta} \right)$

$= \left( \sin{\theta} + \cos{\theta} \right) \left( \cfrac{\sin{\theta}}{\cos{\theta}} + \cfrac{\cos{\theta}}{\sin{\theta}} \right)$

$= \left( \sin{\theta} + \cos{\theta} \right) \left( \cfrac{\sin^{2}{\theta} + \cos^{2}{\theta}}{\sin{\theta} \cos{\theta}} \right)$

$= \left( \sin{\theta} + \cos{\theta} \right) \left( \cfrac{1}{\sin{\theta} \cos{\theta}} \right) \quad \left[\boldsymbol {\because \sin^{2}{\theta} + \cos^{2}{\theta} = 1} \right]$

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

$= \cfrac{1}{\cos{\theta}} + \cfrac{1}{\sin{\theta}}$

$= \sec{\theta} + \csc{\theta}$

$=\text{RHS}$

$\textbf{Hence proved.}$

The value of $tan^2 60^{0}$ is ?

$\textbf{Find the value of}$

$\boldsymbol{\tan^2 60^{0}}$

$\text{We know that}\ \tan 60^{0}=\sqrt{3}$

$\tan^2 60^{0}= \tan 60^{0}* \tan 60^{0}$

$\tan^2 60^{0}=\sqrt{3}*\sqrt{3}$

$\tan^2 60^{0}= 3$

$\textbf{Hence value of }$

$\boldsymbol{\tan^2 60^{0}= 3}$

Evaluate:
$\sin {25^0}\cos {65^0} + \cos {25^0}\;\sin {65^0}$

$= sin 25^0 cos ( 90^0 -25^0 ) + cos 25^0 sin ( 90^0 -25^0)$

$= sin 25^0 sin 25^0 + cos 25^0 cos 25^0$

$= sin^2 25^0 + cos^2 25^0$

$= 1.\,\,\,\,\,\,\,[ \because cos^2 \theta + sin^2 \theta = 1 ]$

If $A$ and $B$ are acute angles such that $\tan A = \dfrac {1}{3}$ , $\tan B = \dfrac{1}{2}$ and $\tan(A + B) = \dfrac {\tan A + \tan B}{1 - \tan A \tan B}$ , show that $A + B = 45^{\circ}$ .

Given,

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

$\tan (A + B) = \dfrac {\left (\dfrac {1}{3} + \dfrac {1}{2}\right )}{1 - \dfrac {1}{3}\times \dfrac {1}{2}} \left [\because \tan A = \dfrac {1}{3}, \tan B = \dfrac {1}{2}\right ]$

$= \dfrac{\dfrac 56}{ \dfrac 5 6}\\$

$= 1$
This implies,

$\tan (A + B) = 1$

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

$A + B = 45^{\circ}$ . Proved

If in $\Delta A B C , \angle B = 90 ^ { \circ } , a = 12 \mathrm { cm } , b = 13 \mathrm { cm } ,$ then write all the trigonometric ratios for $\angle A$ and $\angle C$

Given that

$\angle B = 90^{\circ} , AC = 13\ cm , BC = 12\ cm$

$\because$

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

$\Rightarrow$

$AB = \sqrt{AC^{2} - BC^{2}}$

$= \sqrt{(13)^{2} - (12)^{2}}$

$= \sqrt{169 - 144} = \sqrt{25}$

$= 5\ cm$
For acute angle

$A$ , the value of trigonometric ratios are

$\sin A = \dfrac{BC}{AC} = \dfrac{12}{13} , \cos A = \dfrac{AB}{AC} = \dfrac{5}{13}$ ,

$\tan A = \dfrac{BC}{AB} = \dfrac{12}{5}$

$cosec A = \dfrac{AC}{BC} = \dfrac{13}{12} , \sec A = \dfrac{AC}{AB} = \dfrac{13}{5}$

$\cot A = \dfrac{AB}{BC} = \dfrac{5}{12}$
1869952_1490531_ans_d861507865314cca9b0e3917264b65c9.png

$4 \tan 60 ^ { \circ } \sec 30 ^ { \circ } + \dfrac { \sin 31 ^ { \circ } \sec 59 ^ { \circ } + \cot 59 ^ { \circ } \cot 31 ^ { \circ } } { 8 \sin ^ { 2 } 30 ^ { \circ } - \tan ^ { 2 } 45 ^ { \circ } }$

1326812_1504065_ans_340d71493e7746adbf382657e086f41e.jpg

In $\Delta { ABC } ,$ if $\dfrac { \cos { A } } { a } = \dfrac { \cos { B } } { b },$ then show that it is an isosceles triangle.

$\dfrac { \cos { A } } { a } = \dfrac { \cos { B } } { b },$ ....(i)

$\dfrac{\sin A }{a} = \dfrac{\sin B}{b}$ .........(ii) (True for any triangle)

Hence

$\dfrac{(2)}{(1)}$

$\tan A = \tan B$

$A = B$
So ,

$\Delta ABC$ is isosceles as two angles are equal.

If the angles $(2x-10)^{\circ}$ and $(x-5)^{\circ}$ are complementary angles, find $x (in \, degrees).$

Given

$(2 x-10)^{\circ}$ and

$(x-5)^{\circ}$ are complementary angles

$\implies (2 x-10)^{\circ}+(x-5)^{\circ}=90^{\circ}$

$\implies 3 x-15^{\circ}=90^{\circ}$

$\implies 3 x=105^{\circ}$

$\implies x=35^{\circ}$

If $\tan \theta = \dfrac { 5 } { 4 },$ then find the value of $\dfrac { 3 \sin \theta + 4 \cos \theta } { 3 \sin \theta - 4 \cos \theta }.$

1319581_1539368_ans_a1f46eb0785348e4ac2066f13c0ac0fd.jpeg

If sec $\theta = 5/4$ , show that $\left( \dfrac { sin\theta -2cos\theta }{ tan\theta -cos\theta } \right) = \dfrac{12}{7}$

Given: sec

$\theta = \dfrac{5}{4}$

$sec \theta = \dfrac{5k }{ (4k)} = \dfrac{5}{4}$ and

$cos \theta = \dfrac{4}{5}$

Where k is any positive.

By pythagoras theorem:

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

$BC^2 = AC^2 - AB^2$

$= 25k^2 + 16k^2$

$BC = 3k$

Find

$sin \theta,tan \theta$ and

$cot \theta$ using their definitions:

$sin \theta = \dfrac{ perpendicular}{hypotenuse }= \dfrac{3}{5}$

$tan \theta = \dfrac{perpendicular}{base} = \dfrac{3}{4}$

$cot \theta =\dfrac{ 1 }{tan \theta} = \dfrac{4}{3}$

Now,

LHS

$= \left( \dfrac { sin\theta -2cos\theta }{ tan\theta -cot\theta } \right)$

$= \dfrac { \frac { 3 }{ 5 } -2\times \frac { 4 }{ 5 } }{ \frac { 3 }{ 4 } -\frac { 4 }{ 3 } }$

$LHS= \dfrac { \frac { 3-8 }{ 5 } }{ \frac { 9-16 }{ 12 } }$

$LHS= \dfrac{12}{7}$

$LHS= RHS$

1537561_1713790_ans_3d05c4b318af4dd2932a8cab1434cd2f.png

If tan $\theta = \dfrac{20}{21}$ ,show that $\left( \dfrac { 1-sin\theta +cos\theta }{ 1+sin\theta +cos\theta } \right) = \dfrac{3}{7}$

Given : tan

$\theta = \dfrac{20}{21}$

$tan \theta = \dfrac{20 k}{21k}$

Where k is any positive.

By pythagoras theorem.

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

$441k^2 + 400 k^2$

$AC = 29 k$

Find

$sin \theta$ and

$cos \theta$ using their definitions.

$sin \theta = \dfrac{perpendicular }{ hypotenuse} = \dfrac{20}{29 }$

$cos \theta = \dfrac{ base}{hypotenuse} = \dfrac{21}{29}$

Now,
LHS

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

$= \dfrac { 1-\frac { 20 }{ 20 } +\frac { 21 }{ 20 } }{ 1+\frac { 20 }{ 20 } +\frac { 21 }{ 20 } }$

$LHS= \dfrac{30}{70}$

$LHS=\dfrac{ 3}{7}$

$LHS=RHS$

Hence proved

1537559_1713783_ans_6575b3cb18714233bf3484075f12406b.png

If tan $\theta = \dfrac{1}{ \sqrt 7}$ show that $\dfrac{(cosec^2\theta - sec^2 \theta)}{cosec^2 \theta + sec^2 \theta)} = \dfrac{3}{4}$

Given : tan

$\theta =\dfrac{ 1} {\sqrt 7}$

$tan \theta =\dfrac{k}{(k\surd 7)} = \dfrac{BC}{AC}$

Where

$k$ is any positive

By Pythagoras theorem:

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

$AB^2= K^2 + 7k^2$

$AB = 2 \sqrt 2 k$

Find

$cosec \theta$ and

$sec \theta$ using their definitions:

$cosec \theta = \dfrac{AB}{BC} = 2 \sqrt 2 k$

$sec \theta =\dfrac{AB}{AC} = \dfrac{ 2 \surd 2} { \surd 7 }$

Now,

LHS

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

$= \dfrac { \left( 2\sqrt { 2 } \right) ^ 2-\left( \frac { 2\surd 2 }{ \surd 7 } \right) ^ 2 }{ \left( 2\sqrt { 2 } \right) ^ 2+\left( \frac { 2\surd 2 }{ \surd 7 } \right) ^ 2 }$

$= 48/64$

$LHS=3/4$

$LHS= RHS$

Hence, proved

In a right $\triangle ABC$ , right-angled at $B$ , if $\tan A = 1,$ then verify that, $2 sin A. cos A = 1$

Consider

$\triangle ABC$ to be right angled triangle at

$B.$
angle

$C = 90$ degree

Given :

$tan A = 1$ ...(1)

$tan A = 1 = \dfrac{BC}{AB}$

$AB = BC$

Again,

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

$sin A = cos A$ .... using (1)

By Pythagoras theorem:

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

$AC^2 = 2BC^2$

$\left(\dfrac{AC}{BC}\right)^2 = 2\\$

$\dfrac{AC}{BC} =$

$\sqrt 2$

$cosec\ A$

$= \sqrt 2$

$\sin A = \dfrac 1{\sqrt2}$

$\cos A = \dfrac 1{\sqrt2}$

Now,

$2 \sin A \cos A$

$=2 \times \dfrac 1{\sqrt2} \times \dfrac 1 { \sqrt2 }$

$= 2\times \dfrac 12$

$= 1$
= RHS

1537981_1713929_ans_7c58f1b8bf6e4c419bbe62d201ffa27a.png

In a $\triangle ABC$ angle B = 90 degree, AB = 12 cm and BC = 5 cm
Find (i) cos A (ii) cosec A (iii) cos C (iv) cosec C

From figure:

$\triangle ABC is$ a right angled triangle

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

$AC^2 = (5)^2 + (12)^2$

$AC^2 = 25 + 144$

$AC^2 = 169 = (13)^2$

$AC = 13$

Now from figure,

$i. \cos A = \dfrac{AB}{AC} =\dfrac{ 12}{13}$

$ii. \csc A = \dfrac{AC}{BC} = \dfrac{13}{5}$

$iii. \cos C = \dfrac{BC}{AC} = \dfrac{5}{13}$

$iv. \csc C = \dfrac{AC}{AB} = \dfrac{13}{12}$

Hence, proved.

1537898_1713892_ans_18e0306871124a799a6dbfc0add9b76b.png

If 3 cot $\theta = 2$ ,show that $\left( \dfrac { 4sin\theta -3cos\theta }{ 2sin\theta +6cos\theta } \right) = \dfrac{1}{3}$

Given :

$3 \cot \theta = 2$
or

$\cot \theta = \dfrac 23$
Where

$k$ is any positive.

By Pythagoras theorem:

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

$AC^2 = 2^2 + 3^2$

$AC =$

$\sqrt {13}$

Find

$\sin \theta$ and

$\cos \theta$ using their definitions:

$\cos \theta = \dfrac{perpendicular }{hypotenuse} = \dfrac 3 { \sqrt {13}}$

$\sin \theta = \dfrac{base }{hypotenuse} = \dfrac 2 { \sqrt {13}}$

Now,
LHS

$= \left( \dfrac { 4sin\theta -cos\theta }{ 2sin\theta +6cos\theta } \right) = \dfrac { 4\times \dfrac { 3 }{ \sqrt { 13 } } -3\times \dfrac { 2 }{ \sqrt { 13 } } }{ 2\times \dfrac { 3 }{ \sqrt { 13 } } +6\times \dfrac { 2 }{ \sqrt { 13 } } }$

$= \dfrac { \dfrac { 12-6 }{ \sqrt { 13 } } }{ \dfrac { 6+12 }{ \sqrt { 13 } } }$

$= \dfrac 13$
= RHS

1537594_1713852_ans_e8dc92f243764701b0d4bd4e47f62f4d.PNG

If cot $\theta = 3/4,$ show that
$\left( \frac { sec\theta -cos\theta }{ sec\theta +cosec\theta } \right) = \frac{1}{\sqrt{7}}$

1961089_1713800_ans_e09fe7eaa84747498ef2b05a1ea2df96.jpg

If sin $\theta = \dfrac34,$ show that $\sqrt { \dfrac { cosec^ 2\theta -cot^ 2\theta }{ sec^ 2\theta -1 } } = \dfrac{\sqrt{7}}{3}$ .

Given: sin

$\theta = \dfrac34$
or

$sin \theta = \dfrac{3k}{4k}$
Where

$k$ is any positive.

By pythagoras theorem.

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

$16k^2 = AB^2 + 9k^2$

$AB = \sqrt 7$

Find

$sec \theta$ and

$cot \theta$ using their definitions:

$sec \theta = \dfrac{hypotenuse}{base} = \dfrac4{\sqrt 7}$

$cot \theta = \dfrac{base}{perpendicular} = \dfrac{\sqrt7} 3$

Now,
LHS =

$\sqrt { \dfrac { cosec^ 2\theta -cot^ 2\theta }{ sec^ 2\theta -1 } }$

$= \sqrt { \dfrac { \left( \dfrac { 4 }{ 3 } \right) ^ 2-\left( \dfrac { \sqrt { 7 } }{ 3 } \right) ^ 2 }{ \left( \dfrac { 4 }{ \sqrt { 7 } } \right) ^ 2-1 } }$

$= \sqrt { \dfrac { \dfrac{16}9-\dfrac79 }{ { \dfrac{16}7-1 } } }$

$= \dfrac { \sqrt { 7 } }{ 3 }$
= RHS

1537572_1713810_ans_6a0433271b3c443b8fc8915faa0a6273.png

If $cos \theta = \dfrac{3}{5}$ , show that $\dfrac { \left( sin\theta -cot\theta \right) }{ 2tan\theta } = \dfrac{3}{160}$

Given :

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

$\cos \theta = \dfrac{(3k)}{(5k)} = \dfrac{AC}{AB}$

Where k is any positivity

By Pythagoras theorem:

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

$BC = 4k$

Find

$sin \theta \ \ tan \theta \ \ and \ \ cot \theta$ using their definitions:

$sin \theta = \dfrac{perpendicular}{hypotenuse }= \dfrac{4}{5}$

$tan \theta = \dfrac{perpendicular}{base} = \dfrac{4}{3}$

$cot \theta = \dfrac{ 1}{tan \theta }= \dfrac{3}{4}$

$cot \theta = \dfrac{1}{tan \theta} = \dfrac{3}{4}$

Now

LHS=

$\dfrac { \left( sin\theta -cot\theta \right) }{ 2tan\theta }$

$\dfrac{4/5-3/4}{2(4/3)}$

$= \dfrac{3}{160}$

$LHS=RHS$

1537580_1713825_ans_e84f38ec813d423b86ada160814eadb4.png

If $tan \theta = 4/3$ , show that $(sin \theta + cos \theta) = 7/5$

Given,

$tanθ = 4/3 .$

$cotθ = 1/ ( 4/3 ) = 3/4$

[ Since,

$1/ tanθ = cotθ$ ]

We know that,

$tanθ =$ opposite side to theta / adjacent side to theta.

So, opposite side

$= 4$ , Adjacent side

$= 3$

Hypotenuse =

$\sqrt{(4²+3² )}= 5$

Now,

$sinθ =$ Opposite side to theta / Hypotenuse

$= 4/5$

$cosθ =$ Adjacent side to theta / Hypotenuse

$= 3/5$

To prove :

$sinθ + cosθ = 7/5 .$

L. H.S

$= sinθ + cosθ$

$= 4/5 + 3/5$

$= 7/5$

= R. H. S

Hence proved!

In a $\triangle ABC$ , in which $\angle = 90^o , \angle ABC = \theta^o, BC = 21$ units, $AB = 29$ units, show that $(cos^2 \theta - sin^2 \theta ) = \dfrac{41}{841}$

Draw a triangle using given instructions:
From figure :

$\triangle ABC$ is a right angled triangle

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

$(29)^2 = AC^2 + (21)^2$

$841 = AC^2 + 441$

$AC^2 = 400$
or

$AC = 20$

Find

$\sin \theta$ and

$\cos \theta$

$sin \theta = \dfrac{AC}{AB} = \dfrac{20}{29}\\$

$cos \theta = \dfrac{BC}{ AB} = \dfrac{21}{29}$

Now :
LHS =

$cos^2\theta - sin^2 \theta$

$= {\left(\dfrac{21}{29}\right)}^2 - {\left(\dfrac{20}{29}\right)}^2\\$

$= \dfrac{41}{841}$
= RHS
Hence proved.

1537892_1713887_ans_6215660249ab44d9bbaa4282818b8001.png

If sec $\theta = \dfrac{17}{8},$ verify that $\dfrac{3-4 sin^2 \theta}{4 cos^2 \theta - 3} = \dfrac{3- tan^2 \theta}{1 - 3tan^2 \theta}$

Given :

$sec \theta = 17/8$
or

$sec \theta = 17 k/8 k$

Where k is any positive.
By pythagoras theorem:

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

$BC^2= AC^2 - AB^2$

$BC^2= 289k^2 - 64k^2$

$BC^2=225 k^2$

$BC = 15 k$

Find sin

$\theta$ and

$cos \theta$ using their definition:

$sin \theta =$

$\dfrac{perpendicular}{hypotenuse} = \dfrac{15}7$

$cos \theta$

$= \dfrac{base }{hypotenuse} =\dfrac{ 15}8$

Now,
LHS

$\dfrac{ 3-4 sin^ 2 \theta }{ 4cos^ 2\theta -3 }$

$\dfrac { 3-4\times \left( \dfrac { 15 }{ 17 } \right) ^ 2 }{ 4\times \left( \dfrac { 8 }{ 17 } \right) ^ 2-3 } =\dfrac { 3-\dfrac { 4\times 225 }{ 289 } }{ 4\times \dfrac { 64 }{ 289 } -3 }$

$= \dfrac{(867-900)}{(256-867)}$

$= \dfrac{33}{611}$

RHS:

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

$\dfrac { 3-\dfrac { 225 }{ 64 } }{ 1-3\times \dfrac { 225 }{ 64 } } =\dfrac { \dfrac { 3\times 64-225 }{ 64 } }{ \dfrac { 64-3\times 225 }{ 64 } }$

$= \dfrac{(19-225)}{(64-675)}$

$=\dfrac{33}{611}$

LHS = RHS

Verified.

1537846_1713865_ans_58b6bee817f844a4b9879a7ae8f094bc.png

If $3 \cot \theta = 4$ , show that $\dfrac{(1-\tan^2 \theta)}{(1+\tan^2 \theta)}=(\cos^2 \theta - \sin^2 \theta)$ .

Given:

$3 \cot \theta = 4$
or

$\cot \theta = \dfrac {4k} {3k}$
Where

$k$ is any positive.

By pythagoras theorem ;

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

$= 16k^2 + 9k^2$

$AC = 5 k$

Find

$\sin \theta$ and

$\cos \theta$ using their definition

$\cos \theta = \dfrac{perpendicular }{hypotenuse} = \dfrac 4 5$

$\sin \theta = \dfrac{base }{hypotenuse} = \dfrac 35$

Now,
LHS =

$\dfrac{(1-tan^2 \theta )}{(1+tan^2 \theta )} \\$

$= \dfrac { 1-\left( \dfrac 34 \right) ^ 2 }{ 1+\left( \dfrac 3 4 \right) ^ 2 } \\$

$= \dfrac { 1-\dfrac 9{16} }{ 1+\dfrac 9{16} } =\dfrac { 7 }{ 16 } \times \dfrac { 16 }{ 25 }$

$= \dfrac{7}{25}$

Again,
RHS =

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

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

$= \dfrac{16}{25} - \dfrac9{25}\\$

$= \dfrac 7{25}$
Therefore, RHS = LHS
Hence proved.

1537597_1713860_ans_0e92136a2a5540fd9f204a861892266d.png

Without using trigonometric tables, prove that:
$\tan ^{ 2 }{ { 66 }^{ o } } -\cot ^{ 2 }{ { 24 }^{ o } } =0$

$LHS=\tan ^{ 2 }{ { 66 }^{ o } } -\cot ^{ 2 }{ { 24 }^{ o } } =\tan ^{ 2 }{ { 66 }^{ o } } -\cot ^{ 2 }{ { \left( 90-66 \right) }^{ o } } =\tan ^{ 2 }{ { 66 }^{ o } } -\tan ^{ 2 }{ { 66 }^{ o } } =0=RHS\quad$

Without using trigonometric tables, prove that:
$\left( \sin { { 65 }^{ o } } +\cos { { 25 }^{ o } } \right) \left( \sin { { 65 }^{ o } } -\cos { { 25 }^{ o } } \right) =0$

$LHS=\left( \sin { { 65 }^{ o } } +\cos { { 25 }^{ o } } \right) \left( \sin { { 65 }^{ o } } -\cos { { 25 }^{ o } } \right) =\sin ^{ 2 }{ { 65 }^{ o } } -\cos ^{ 2 }{ { 25 }^{ o } } =\sin ^{ 2 }{ { 65 }^{ o } } -\cos ^{ 2 }{ { \left( 90-65 \right) }^{ o } }$

$=\sin ^{ 2 }{ { 65 }^{ o } } -\sin ^{ 2 }{ { 65 }^{ o } } =0=RHS\quad$

Without trigonometric tables, prove that:
$\cos { { 54 }^{ o } } \cos { { 36 }^{ o } } -\sin { { 54 }^{ o } } \sin { { 36 }^{ o } } =0\quad$

$LHS\\=\cos { { 54 }^{ o } } \cos { { 36 }^{ o } } -\sin { { 54 }^{ o } } \sin { { 36 }^{ o } } \\=\cos { { 54 }^{ o } } \cos { { 36 }^{ o } } -\sin { \left( { 90 }^{ o }-{ 36 }^{ o } \right) } \sin { \left( { 90 }^{ o }-{ 36 }^{ o } \right) }$

$\\=\cos { { 54 }^{ o } } \cos { { 36 }^{ o } } -\cos { { 54 }^{ o } } \cos { { 36 }^{ o } } \\=0\\=RHS$

Without using trigonometric tables, prove that:
$\sin ^{ o }{ { 48 }^{ o } } +\sin ^{ o }{ { 42 }^{ o } } =1$

$LHS=\sin ^{ 2 }{ { 48 }^{ o } } +\sin ^{ 2 }{ { 42 }^{ o } } =\sin ^{ 2 }{ { 48 }^{ o } } +\sin ^{ 2 }{ { \left( 90-48 \right) }^{ o } } =\sin ^{ 2 }{ { 48 }^{ o } } +\cos ^{ 2 }{ { 48 }^{ o } } =1=RHS$

Without using trigonometric tables, prove that:
$co\sec ^{ 2 }{ { 72 }^{ o } } -\tan ^{ 2 }{ { 18 }^{ o } } =1$

$LHS=co\sec ^{ 2 }{ { 72 }^{ o } } -\tan ^{ 2 }{ { 18 }^{ o } } =co\sec ^{ 2 }{ { 72 }^{ o } } -\tan ^{ 2 }{ { \left( 90-72 \right) }^{ o } } =co\sec ^{ 2 }{ { 72 }^{ o } } -\cot ^{ 2 }{ 72 } =1=RHS$

Without using trigonometric tables, prove that:
$\cos ^{ 2 }{ { 57 }^{ o } } -\sin ^{ 2 }{ { 33 }^{ o } } =0$

$LHS=\cos ^{ 2 }{ { 57 }^{ o } } -\sin ^{ 2 }{ { 33 }^{ o } } =\cos ^{ 2 }{ { 57 }^{ o } } -\sin ^{ 2 }{ { \left( 90-57 \right) }^{ o } } =\cos ^{ 2 }{ { 57 }^{ o } } -\cos ^{ 2 }{ { 57 }^{ o } } =0=RHS$

Without trigonometric tables, prove that:
$\sin { { 53 }^{ o } } \cos { { 37 }^{ o } } +\cos { { 53 }^{ o } } \sin { { 37 }^{ o } } =1$

$LHS\\=\sin { { 53 }^{ o } } \cos { { 37 }^{ o } } +\cos { { 53 }^{ o } } \sin { { 37 }^{ o } } \\=\sin { { 53 }^{ o } } \cos { { \left( 90-53 \right) }^{ o } } +\cos { { 53 }^{ o } } \sin { { \left( 90-53 \right) }^{ o } }$

$\\=\sin { { 53 }^{ o } } \times \sin { { 53 }^{ o } } +\cos { { 53 }^{ o } } \times \cos { { 53 }^{ o } } \\=\sin ^{ 2 }{ { 53 }^{ o } } +\cos ^{ 2 }{ { 53 }^{ o } } \\=1\\=RHS\quad$

Without trigonometric tables, prove that:
$\sec { { 70 }^{ o } } \sin { { 20 }^{ o } } +\cos { { 20 }^{ o } } co\sec { { 70 }^{ o } } =2$

$LHS\\=\sec { { 70 }^{ o } } \sin { { 20 }^{ o } } +\cos { { 20 }^{ o } } co\sec { { 70 }^{ o } } \\=\sec { \left( { 90 }^{ o }-{ 20 }^{ o } \right) } \sin { { 20 }^{ o } } +\cos { { 20 }^{ o } } co\sec { \left( { 90 }^{ o }-{ 20 }^{ o } \right) }$

$\\=co\sec { { 20 }^{ o } } \sin { { 20 }^{ o } } +\cos { { 20 }^{ o } } \sec { { 20 }^{ o } } \\=1+1\\=2\\=RHS$

Without using trigonometric tables, prove that:
$\cos ^{ 2 }{ { 75 }^{ o } } +\cos ^{ 2 }{ { 15 }^{ o } }=1$

$LHS=\cos ^{ 2 }{ { 75 }^{ o } } +\cos ^{ 2 }{ { 15 }^{ o } } =\cos ^{ 2 }{ { 75 }^{ o } } +\sin ^{ 2 }{ { \left( 90-15 \right) }^{ o } } =\cos ^{ 2 }{ { 75 }^{ o } } +\sin ^{ 2 }{ { 75 }^{ o } } =1=RHS$

Without trigonometric tables, prove that:
$\sin { { 35 }^{ o } } \sin { { 55 }^{ o } } -\cos { { 35 }^{ o } } \cos { { 55 }^{ o } } =0$

$LHS\\=\sin { { 35 }^{ o } } \sin { { 55 }^{ o } } -\cos { { 35 }^{ o } } \cos { { 55 }^{ o } } \\=\sin { \left( { 90 }^{ o }-{ 55 }^{ o } \right) } \sin { \left( { 90 }^{ o }-{ 55 }^{ o } \right) } -\cos { { 35 }^{ o } } \cos { { 55 }^{ o } }$

$\\=\cos { { 55 }^{ o } } \cos { { 55 }^{ o } } -\cos { { 55 }^{ o } } \cos { { 55 }^{ o } } \\=0\\=RHS\quad$

Without trigonometric tables, prove that:
$\left( \sin { { 72 }^{ o } } +\cos { { 18 }^{ o } } \right) \left( \sin { { 72 }^{ o } } -\cos { { 18 }^{ o } } \right) =0$

$LHS\\=\left( \sin { { 72 }^{ o } } +\cos { { 18 }^{ o } } \right) \left( \sin { { 72 }^{ o } } -\cos { { 18 }^{ o } } \right) \\=\sin ^{ 2 }{ { 72 }^{ o } } -\cos ^{ 2 }{ \left( { 90 }^{ o }-{ 55 }^{ o } \right) }$

$\\=\sin ^{ 2 }{ { 72 }^{ o } } -\sin ^{ 2 }{ { 72 }^{ o } } \\=0\\=RHS$

Prove that:
$\sec^{2}\theta \textrm{cosec}^{2}\theta = \tan^{2}\theta + \cot^{2}\theta + 2$

1557364_1730754_ans_fbb653eca7bb441e9e03dd7692408e0d.PNG

Prove the following statement:
$(\sec A + \cos A) (\sec A - \cos A) = \tan^{2}A + \sin^{2}A$ .

1565302_1730659_ans_22b464e082654e7f81fdeec380a98f58.png

Prove that
$\cfrac { \cos { { 80 }^{ o } } }{ \sin { { 10 }^{ o } } } +\cos { { 59 }^{ o } } co\sec { { 31 }^{ o } } =2$

$\quad LHS\\=\cfrac { \cos { { 80 }^{ o } } }{ \sin { { 10 }^{ o } } } +\cos { { 59 }^{ o } } co\sec { { 31 }^{ o } } \\=\cfrac { \cos { { 80 }^{ o } } }{ \sin { \left( { 90 }^{ o }-{ 59 }^{ o } \right) } } +\cos { { 59 }^{ o } } co\sec { \left( { 90 }^{ o }-{ 59 }^{ o } \right) } \\=1+1\\=2\\=RHS$

Prove the following statements.
$\dfrac {1 + \tan^{2}A}{1 + \cot^{2}A} = \dfrac {\sin^{2}A}{\cos^{2}A}$ .

1565326_1730732_ans_7301aa71ae3f4212a730928e0bcc1cad.png

Prove that,
$\dfrac {\sec A - \tan A}{\sec A + \tan A} = 1 - 2\sec A \tan A + 2\tan^{2}A$ .

1591151_1730734_ans_268e052fb45d41ad97adb4474ffaea80.jpg

Without trigonometric tables, prove that:
$\tan { { 48 }^{ o } } \tan { { 23 }^{ o } } \tan { { 42 }^{ o } } \tan { { 67 }^{ o } } =1$

$LHS\\=\tan { { 48 }^{ o } } \tan { { 23 }^{ o } } \tan { { 42 }^{ o } } \tan { { 67 }^{ o } } \\=\tan { { 48 }^{ o } } \tan { { 23 }^{ o } } \tan { \left( { 90 }^{ o }-{ 48 }^{ o } \right) } \tan { \left( { 90 }^{ o }-{ 48 }^{ o } \right) }$

$\\=\tan { { 48 }^{ o } } \tan { { 23 }^{ o } } \cot { { 48 }^{ o } } \cot { { 23 }^{ o } } \\=1\times 1\\=1\\=RHS$

Prove that
$\cfrac { \sin { { 70 }^{ o } } }{ \cos { { 20 }^{ o } } } +\cfrac { co\sec { { 20 }^{ o } } }{ \sec { { 70 }^{ o } } } -2\cos { { 70 }^{ o } } co\sec { { 20 }^{ o } } =0$

$LHS\\=\cfrac { \sin { { 70 }^{ o } } }{ \cos { { 20 }^{ o } } } +\cfrac { co\sec { { 20 }^{ o } } }{ \sec { { 70 }^{ o } } } -2\cos { { 70 }^{ o } } co\sec { { 20 }^{ o } }$

$\\=\cfrac { \sin { { 70 }^{ o } } }{ \cos { \left( { 90 }^{ o }-{ 70 }^{ o } \right) } } +\cfrac { co\sec { \left( { 90 }^{ o }-{ 70 }^{ o } \right) } }{ \sec { { 70 }^{ o } } } -2\cos { { 70 }^{ o } } co\sec { \left( { 90 }^{ o }-{ 70 }^{ o } \right) }$

$\\=\cfrac { \sin { { 70 }^{ o } } }{ \sin { { 70 }^{ o } } } +\cfrac { \sec { { 70 }^{ o } } }{ \sec { { 70 }^{ o } } } -2\cos { { 70 }^{ o } } \sec { { 70 }^{ o } } \\=1+1-2\\=0\\=RHS$

Prove that
$\cfrac { 2\sin { { 68 }^{ o } } }{ \cos { { 22 }^{ o } } } -\cfrac { 2\cot { { 15 }^{ o } } }{ 5\tan { { 75 }^{ o } } } -\cfrac { 3\tan { { 45 }^{ o } } \tan { { 20 }^{ o } } \tan { { 40 }^{ o } } \tan { { 50 }^{ o } } \tan { { 70 }^{ o } } }{ 5 } =1\quad$

$LHS\\=\cfrac { 2\sin { { 68 }^{ o } } }{ \cos { { 22 }^{ o } } } +\cfrac { 2\cot { { 15 }^{ o } } }{ 5\tan { { 75 }^{ o } } } -\cfrac { 3\tan { { 45 }^{ o } } \tan { { 20 }^{ o } } \tan { { 40 }^{ o } } \tan { { 50 }^{ o } } \tan { { 70 }^{ o } } }{ 5 }$

$\\=\cfrac { 2\sin { { 68 }^{ o } } }{ \cos { \left( { 90 }^{ o }-{ 68 }^{ o } \right) } } +\cfrac { 2\cot { { 15 }^{ o } } }{ 5\tan { \left( { 90 }^{ o }-{ 15 }^{ o } \right) } } -\cfrac { 3\tan { { 45 }^{ o } } \tan { { 20 }^{ o } } \tan { { 40 }^{ o } } \tan { \left( { 90 }^{ o }-{ 50 }^{ o } \right) } \tan { \left( { 90 }^{ o }-{ 20 }^{ o } \right) } }{ 5 }$

$\\=\cfrac { 2\sin { { 68 }^{ o } } }{ \sin { { 68 }^{ o } } } +\cfrac { 2\cot { { 15 }^{ o } } }{ 5\cot { { 15 }^{ o } } } -\cfrac { 3\tan { { 45 }^{ o } } \tan { { 20 }^{ o } } \tan { { 40 }^{ o } } \cot { { 40 }^{ o } } \cot { { 20 }^{ o } } }{ 5 } \\=2-(2/5)-(3/5)\\=(10-2-3)/5\\=1\\=RHS$

Prove the following statements.
$\sqrt {co\sec^{2}A - 1} = \cos A\cdot \csc A$ .

1557247_1730751_ans_96973399501b4147b2f66e30a3320b30.jpg

Prove the following statements.
$\tan^{2}A - \sin^{2}A = \sin^{4} A \sec^{2}A$ .

1557367_1730756_ans_95310b95a4dc4065ab0c12aaf3bc8b28.jpg

Prove that following identities, where the angles involved are acute angles for which the trigonometric ratios as defined: $\left ( 1-\tan ^{2}A \right )/\left ( \cot ^{2}A-1 \right )=\tan ^{2}A$ .

$L.H.S.=\frac{1-\tan ^{2}A}{\cot^{2}A-1 }=\frac{1-\frac{\sin ^{2}A}{\cos ^{2}A}}{\frac{\cos ^{2}A}{\sin ^{2}A}-1}$

$\frac{\frac{\cos ^{2}A-\sin ^{2}A}{\cos ^{2}A}}{\frac{\cos ^{2}A-\sin ^{2}A}{\sin ^{2}A}}$

$\frac{\cos ^{2}A-\sin ^{2}A}{\cos ^{2}A}*\frac{\sin ^{2}A}{\cos ^{2}A-\sin ^{2}A}$

$\frac{\sin ^{2}A}{\cos ^{2}A}=\tan ^{2}A=R.H.S.$

If $\tan A=1/\sqrt{3}$ , find all other trigonometric ratios of angle $A$ .

Given,

$\tan A=1/\sqrt{3}$ ,

In right

$\Delta ABC$

$\tan A=BC/AB=1/\sqrt{3}$ ,

$BC=1$ and

$AB=\sqrt{3}$ .

By Pythagoras theorem,

$AC=\sqrt{\left ( AB^{2}+BC^{2} \right )}$

$\sqrt{\left ( \sqrt{3} \right )^{2}+\left ( 1 \right )^{2}}$

$\sqrt{\left ( 3+1 \right )}$

$\sqrt{4}=2$ .So,

$\sin A=BC/AC=1/2$

$\cos A=AB/AC=\sqrt{3}/2$

$\cot A=1/\tan A=\sqrt{3}$

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

$cosecA$ =

$1/\sin A=2/1=2$ .

Prove that following identities, where the angles involved are acute angles for which the trigonometric ratios as defined: $\cos \theta /\left ( 1-\tan \theta \right )-\sin ^{2}\theta /\left ( \cos \theta -\sin \theta \right )=\cos \theta +\sin \theta$ .

$L.H.S.=\frac{\cos \theta }{1-\tan \theta}-\frac{\sin ^{2}\theta}{\cos \theta -\sin \theta}$

$\frac{\cos \theta}{1-\frac{\sin \theta }{\cos \theta}}-\frac{\sin ^{2}\theta}{\cos \theta -\sin \theta}$

$\frac{\cos \theta}{\frac{\cos \theta -\sin \theta}{\cos \theta}}-\frac{\sin ^{2}\theta}{\cos \theta -\sin \theta}$

$\frac{\cos ^{2}\theta}{\cos \theta -\sin \theta}-\frac{\sin ^{2}\theta}{\cos \theta -\sin \theta}$

$\frac{\cos ^{2}\theta -\sin ^{2}\theta}{\cos \theta -\sin \theta}$

$\frac{\left ( \cos \theta +\sin \theta \right )\left ( \cos \theta -\sin \theta \right )}{\cos \theta -\sin \theta}$

$\cos \theta +\sin \theta = R.H.S.$

Prove that:
$\left (\dfrac {1}{\sec^{2}\alpha - \cos^{2}\alpha} + \dfrac {1}{\text{cosec}^{2}\alpha - \sin^{2}\alpha}\right ) \cos^{2} \alpha \sin^{2}\alpha = \dfrac {1 - \cos^{2} \alpha \sin^{2} \alpha}{2 + \cos^{2} \alpha \sin^{2} \alpha}$ .

1591155_1730772_ans_0414658ed11b4b85a4a60054da766e59.jpg

Without using trigonometric tables, evaluate the following: $\left ( \sin 29^{0}/cosec61^{0} \right )+2\cot 8^{0}\cot 17^{0}\cot 45^{0}\cot 73^{0}\cot 82^{0}-3\left ( \sin ^{2}38^{0}+\sin ^{2}52^{2} \right )$ .

$\left ( \sin 29^{0}/cosec61^{0} \right )+2\cot 8^{0}\cot 17^{0}\cot 45^{0}\cot 73^{0}\cot 82^{0}-3\left ( \sin ^{2}38^{0}+\sin ^{2}52^{2} \right )$

$\left ( \sin 29^{0}/cosec\left ( 90^{0}-29^{0} \right ) \right )+\left [ 2\cot 8^{0}\cot 17^{0}\cot 45^{0}\cot \left ( 90^{0}-17^{0} \right )\cot \left ( 90^{0}-8^{0} \right ) \right ]-3\left ( \sin ^{2}38^{2}+\sin ^{2}\left ( 90^{0}-30^{0} \right ) \right )$

$\left ( \sin 29^{0}/\sin 29^{0} \right )+\left [ 2\cot 8^{0}\cot 17^{0}\cot 45^{0}\tan 17^{0}\tan 8^{0} \right ]-3\left ( \sin ^{2}38^{0}+\cos ^{2}38^{0} \right )$

$1+2\left [ \left ( \cot 8^{0}\tan 8^{0} \right )\left ( \cot 17^{0}\tan 17^{0} \right )\cot 45^{0} \right ]-3\left ( 1 \right )$

$1+2\left [ 1*1*1 \right ]-3$

$1+2-3=0$

Without using trigonometric tables, evaluate the following: $\frac{cosec^{2}\left ( 90-\theta \right )-\tan ^{2}\theta }{2\left ( \cos ^{2}48^{0}+\cos ^{2}42^{0} \right )}-\frac{2\tan ^{2}30^{0}\sec ^{2}52^{0}\sin ^{2}38^{0}}{cosec^{2}70^{0}-\tan ^{2}20^{0}}$ .

$\frac{cosec^{2}\left ( 90-\theta \right )-\tan ^{2}\theta }{2\left ( \cos ^{2}48^{0}+\cos ^{2}42^{0} \right )}-\frac{2\tan ^{2}30^{0}\sec ^{2}52^{0}\sin ^{2}38^{0}}{cosec^{2}70^{0}-\tan ^{2}20^{0}}$

$\frac{\sec ^{2}\theta -\tan ^{2}\theta }{2\left ( \cos ^{2}48^{0}+\cos ^{2}\left ( 90^{0}-48^{0} \right ) \right )}-\frac{2\left [ \tan ^{2}30^{0}\sec ^{2}52^{0}\sin ^{2}\left ( 90^{0}-52^{0} \right ) \right ]}{cosec^{2}70^{0}-\tan ^{2}\left ( 90^{0}-70^{0} \right )}$

$\frac{1}{2\left [ \cos ^{2}48^{0}+\sin ^{2}48^{0} \right ]}-\frac{2\left [ \left ( \frac{1}{\sqrt{3}} \right )^{2}\sec ^{2}52^{0}\cos ^{2}52^{0} \right ]}{cosec^{2}70^{0}-\cot ^{2}70^{0}}$

$\frac{1}{2*1}-\frac{2\left [ \frac{1}{3}*1 \right ]}{1}$

$\frac{1}{2}-\frac{2}{3}=\frac{3-4}{6}=\frac{-1}{6}$ .

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

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

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

Show that
$\dfrac {4}{3} \cot^{2} 30^{\circ} + 3\sin^{2} 60^{\circ} - 2cosec^{2} 60^{\circ} - \dfrac {3}{4} \tan^{2} 30^{\circ} = 3\dfrac {1}{3}$ .

1687997_1731226_ans_bef6ad0ef5bc407cb11ae26e5ad84c48.jpg

Verify that
$\sin^{2} 30^{\circ} + \sin^{2} 45^{\circ} + \sin^{2} 60^{\circ} = \dfrac {3}{2}$ .

1590513_1731216_ans_9bed5c094170408f86a3087494dfb820.jpg

Prove the following statement.
$\dfrac {\cos A cosec A - \sin A \sec A}{\cos A + \sin A} = cosec A - \sec A$ .

1557411_1730775_ans_ff8613005cfa490ab308042d24790223.jpg

Show that
$\tan 48^{\circ} \tan 16^{\circ} \tan 42^{\circ} \tan 74^{\circ} = 1$

$\tan 48^{\circ} \tan 16^{\circ} \tan 42^{\circ} \tan 74^{\circ} = 1$

$= \tan 48^{\circ} \tan 16^{\circ} \tan (90^{\circ} - 48^{\circ}) \tan (90^{\circ} - 16^{\circ})$

$= \tan 48^{\circ} \tan 16^{\circ} \cot 48^{\circ} \cot 16^{\circ}$

$= \tan 48^{\circ} \cot 48^{\circ} \tan 16^{\circ} \cot 16^{\circ}$

$= \tan 48^{\circ} \dfrac{1}{\tan 18^{\circ}} \tan 16^{\circ} \dfrac{1}{\tan 16^{\circ}} = 1$

Prove the following identities:
$\dfrac{sec A -1}{secA +1} = \dfrac{1-cosA}{1+ cosA}$

$LHS= \dfrac{sec A -1}{sec A +1}= \dfrac{\dfrac{1}{cos A}-1}{\dfrac{1}{cos A}+1}$

$\dfrac{1- cos A}{1+ cos A}= RHS$
-Hence Proved

Show that
$\cos 36^{\circ} \cos 54^{\circ} - \sin 36^{\circ} \sin 54^{\circ} = 0$

$\cos 36^{\circ} \cos 54^{\circ} - \sin 36^{\circ} \sin 54^{\circ} = 0$

$cos (90^{\circ} - 54^{\circ}) cos (90^{\circ} - 36^{\circ}) - \sin 36^{\circ} \sin 54^{\circ}$

$= \sin 54^{\circ} \sin 36^{\circ} - \sin 36^{\circ} \sin 54^{\circ}$

$= \sin 36^{\circ} \sin 54^{\circ} - \sin 36^{\circ} \sin 54^{\circ} = 0$

Prove the following identities:
$\dfrac{1+sinA}{1-sinA}= \dfrac{cosec A + 1}{cosec A -1}$

$LHS= \dfrac{1+ sin A}{1-sin A}$

$RHS = \dfrac{cosec A + 1}{cosec A -1} = \dfrac{\dfrac{1}{sin A} + 1}{\dfrac{1}{sin A} -1}$

$=\dfrac{1+ sin A}{1- sin A}$
-Hence Proved

Evaluate:
$3 \dfrac{sin 72^\circ}{cos 18^\circ}- \dfrac{sec 32^\circ}{cosec58^\circ}$

$3 \dfrac{sin (90-18)^\circ}{cos 18^\circ}- \dfrac{sec (90-58)^\circ}{cosec58^\circ}$

$= 3 \dfrac{cos 18^\circ}{cos 18^\circ} - \dfrac{cosec 58^\circ}{cosec 58^\circ}= 3-1= 2$

Evaluate
$\sin 15^{\circ} \sec 75^{\circ}$

$\sin 15^{\circ} \sec 75^{\circ}$

$= sin 15^{\circ} sec (90^{\circ} - 15^{\circ})$

$= sin 15^{\circ} cosec 15^{\circ}$

$= sin 15^{\circ} \dfrac{1}{sin 15^{\circ}} = 1$

Evaluate
$\tan 26^{\circ} \tan 64^{\circ}$

$\tan 26^{\circ} \tan 64^{\circ}$

$= \tan 26^{\circ} \tan (90^{\circ} - 26^{\circ})$

$= \tan 26^{\circ} \cot 26^{\circ}$

$= \tan 26^{\circ} \times \dfrac{1}{\tan 26^{\circ}} = 1$

If $\sec \theta =\cfrac{13}{5}$ , show that $\cfrac { 2\sin { \theta } -3\cos { \theta } }{ 4\sin { \theta } -9\cos { \theta } } =3$

we know that

$\sec { \theta } =\cfrac { hypotenuse }{ base }$

$\sec { \theta } =\cfrac { 13 }{ 5 } \Rightarrow \cfrac { H }{ B } =\cfrac { 13 }{ 5 } \Rightarrow \cfrac { AC }{ BC } =\cfrac { 13 }{ 5 } \quad \quad$
Let

$BC=5k$ and

$AC=13k$
where

$k$ is any positive integer
In right angled

$\triangle ABC$ we have

$\Rightarrow { \left( AB \right) }^{ 2 }+{ \left( BC \right) }^{ 2 }={ \left( AC \right) }^{ 2 }\Rightarrow { \left( AB \right) }^{ 2 }+{ \left( 5k \right) }^{ 2 }={ \left( 13k \right) }^{ 2 }\Rightarrow { \left( AB \right) }^{ 2 }={ 169k }^{ 2 }+{ 25k }^{ 2 }\quad \quad$ (Pythagoras theorem)

${ \left( AB \right) }^{ 2 }={ 144k }^{ 2 }\Rightarrow AB=\pm 12k$
(taking positive sqaureroot as side cannot be negative)
Now we have to find the other trigonometric ratios

$\sin { \theta } =\cfrac { Perpendicular }{ hypotenuse } =\cfrac { AB }{ AC } =\cfrac { 12k }{ 13k } =\cfrac { 12 }{ 13 } ;\cos { \theta } =\cfrac { base }{ hypotenuse } =\cfrac { BC }{ AC } =\cfrac { 5k }{ 13k } =\cfrac { 5 }{ 13 } \quad$

$LHS=\cfrac { 2\sin { \theta } -3\cos { \theta } }{ 4\sin { \theta } -9\cos { \theta } } =\cfrac { 2\left( \cfrac { 12 }{ 13 } \right) -3\left( \cfrac { 5 }{ 13 } \right) }{ 4\left( \cfrac { 12 }{ 13 } \right) -9\left( \cfrac { 5 }{ 13 } \right) } =\cfrac { 9 }{ 3 } =3=RHS$
1791356_1814191_ans_1c0fc9b183294fac89c867cbf6ee382c.PNG

1791356_1814191_ans_1c0fc9b183294fac89c867cbf6ee382c.PNG

Evaluate
$\dfrac{\tan 36^{\circ}}{\cot 54^{\circ}}$

$\dfrac{\tan 36^{\circ}}{\cot 54^{\circ}} = \dfrac{tan 36^{\circ}}{cot (90^{\circ} - 36^{\circ})} = \dfrac{tan 36^{\circ}}{cot 36^{\circ}} = 1$

$2 \dfrac{tan57^\circ}{cot 33^\circ} - \dfrac{cot 70^\circ}{tan20^\circ} - \sqrt{2} cos 45^\circ$

$2 \dfrac{tan(90-33)^\circ}{cot 33^\circ} - \dfrac{cot (90-20)^\circ}{tan20^\circ} - \sqrt{2} cos 45^\circ$

$= 2 \dfrac{cot 33^\circ}{cot 33^\circ}- \dfrac{tan 20^\circ }{tan 20^\circ}-\sqrt 2\times \dfrac1{\sqrt2}$

$= 2-1-1$

$= 0$

$\dfrac{cot^2 41^\circ}{tan^2 49^\circ} - 2 \dfrac{sin^2 75^\circ}{cos^2 15^\circ}$

$\dfrac{cot^2 (90-49)^\circ}{tan^2 49^\circ} - 2 \dfrac{sin^2 (90-15)^\circ}{cos^2 15^\circ}$

$= \dfrac{tan^2 49^\circ}{tan^2 49^\circ}-2 \dfrac{cos^2 15^\circ}{cos^2 15^\circ}$

$= 1-2= -1$

Find the value of:

$\dfrac{3 sin 72^\circ}{ cos 18^\circ} - \dfrac{sec 32^\circ}{cosec 58^\circ}$

Given:

$\dfrac{3 sin 72^\circ}{ cos 18^\circ} - \dfrac{sec 32^\circ}{cosec 58^\circ}$

It can also be written as:

$=\dfrac{3 sin (90-18)^\circ}{ cos 18^\circ} - \dfrac{sec (90-58)^\circ}{cosec 58^\circ}$

$=\dfrac{3cos 18^\circ}{cos 18^\circ}- \dfrac{cosec 58^\circ}{cosec 58^\circ}$ { Using

$sin(90^o-A)=cosA, sec(90^o-A)=cosecA$ }

$= 3-1= 2$

Find $AD$ (Refer figure)

From triangle

$ABD$

$\sin B = \dfrac {AD}{AB}$

$\sin 30 = \dfrac {AD}{100}$ [Since

$\angle ACD$ is the exterior angle of the triangle

$ABC$ ]

$\dfrac {1}{2} = \dfrac {AD}{100}$

$AD = 50\ m$ .

$\dfrac{5 \ sin 66^\circ}{cos 24^\circ}- \dfrac{2 \ cot 85^\circ}{tan 5^\circ}$

$\dfrac{5 \ sin (90-24)^\circ}{cos 24^\circ}- \dfrac{2 \ cot 90-5)^\circ}{tan 5^\circ}$

$= \dfrac{5 \ cos 24^\circ}{cos 24^\circ} - \dfrac{2 tan 5^\circ}{tan 5^\circ}$

$5-2= 3$

$sin 27^\circ \ sin 63^\circ- cos 63^\circ \ cos 27^\circ$

$= sin (90- 63)^\circ \ sin 63^\circ - cos 63^\circ \ cos (90- 63)^\circ$

$cos 63^\circ \ sin 63^\circ- cos 63^\circ \ sin 63^\circ$

$= 0$

In the given figure, $AB$ and $EC$ are parallel to each other. Sides $AD$ and $BC$ are $2\ cm$ each and are perpendicular to $AB$ .
Given that $\angle AED = 60^{\circ}$ and $\angle ACD = 45^{\circ}$ . Calculate $AB$ .

From the triangle

$ADC$ we have

$\tan 45^{\circ} = \dfrac {AD}{DC}$

$1 = \dfrac {2}{DC}$

$DC = 2$

$ABCD$ is a parallelogram as

$AD\parallel DC$ and

$AD\perp EC$

Hence, opposite sides are equal.

$AB = DC = 2\ cm$ .

Find the length of $AB$ .

From triangle

$ADE$

$\tan 45^{\circ} = \dfrac {AE}{DE}$

$1 = \dfrac {AE}{30}$

$AE = 30\ cm$

From triangle

$DBE$

$\tan 60^{\circ} = \dfrac {BE}{DE}$

$\sqrt {3} = \dfrac {BE}{30}$

$BE = 30\sqrt {3} cm$

$AB = AE + BE = 30 + 30\sqrt {3} = 30(1 + \sqrt {3}) cm$ .

Evaluate:
$sec 26^\circ \ sin 64^\circ + \dfrac{cosec 33^\circ}{sec 57^\circ}$

$sec 26^\circ \ sin 64^\circ + \dfrac{cosec 33^\circ}{sec 57^\circ}$

$sec (90-64)^\circ \ sin 64^\circ + \dfrac{cosec (90-57)^\circ}{sec 57^\circ}$

$= cosec 64^\circ \ sin 64^\circ + \dfrac{sec 57^\circ}{sec 57^\circ}$

$= 1+1= 2$

Evaluate:
$2 \left( \dfrac{tan 35^\circ}{cot 55^\circ} \right)^2+ \left(\dfrac{cot 55^\circ}{tan 35^\circ} \right)^2 - 3 \left( \dfrac{sec 40^\circ}{cosec 50^\circ} \right)$

$2 \left( \dfrac{tan 35^\circ}{cot 55^\circ} \right)^2+ \left(\dfrac{cot 55^\circ}{tan 35^\circ} \right)^2 - 3 \left( \dfrac{sec 40^\circ}{cosec 50^\circ} \right)$

$2 \left( \dfrac{tan (90-55)^\circ}{cot 55^\circ} \right)^2+ \left(\dfrac{cot (90-35)^\circ}{tan 35^\circ} \right)^2 - 3 \left( \dfrac{sec (90-50)^\circ}{cosec 50^\circ} \right)$

$= 2 \left(\dfrac{cot 55^\circ}{cot 55^\circ} \right)^2 + \left( \dfrac{tan 35^\circ}{tan 35^\circ} \right)^2 - 3 \left(\dfrac{cosec 50^\circ}{cosec 50^\circ} \right)$

$= 0$

Find $AD$ .(Refer figure)

From right triangle

$ABE$

$\tan 45^{\circ} = \dfrac {AE}{BE}$

$1 = \dfrac {AE}{BE}$

$AE = BE$

$AE = BE = 50\ m$ .

From the rectangle

$BCDE$

$DE = BC = 10\ m$ .

The length of

$AD$ will be:

$AD = AE + DE = 50 + 10 = 60\ m$ .

Evaluate: $\dfrac{\cos 22^{0}}{\sin68^{0}}$

$\dfrac{\cos22^{0}}{\sin68^{0}}\\=\dfrac{\cos(90^{0}-68^{0})}{\sin68^{0}}\\=\dfrac{\sin68^{0}}{\sin68^{0}}\\=1$

Find $AD$ .(Refer figure)

From right angled triangle

$ABD$

$\cos A = \dfrac {AD}{AB}$

$\cos 60^{\circ} = \dfrac {AD}{AB}$

$\dfrac {1}{2} = \dfrac {AD}{12}$

$AD = \dfrac {12}{2}$

$= 6\ cm$ .

Find $BC$ .(Refer figure)

From right angled triangle

$ABC$

$\tan 30^{\circ} = \dfrac {AB}{BC}$

$\dfrac {1}{\sqrt {3}} = \dfrac {12}{BC}$

$BC = 12\sqrt {3} cm$ .

Refer given figure, Find $AB$ and $BC$

Let

$BC = xm$

$BD = BC + CD = (x + 20) cm$

In triangle

$ABD$ ,

$\tan 30^{\circ} = \dfrac {AB}{BD}$

$\dfrac {1}{\sqrt {3}} = \dfrac {AB}{BD}$

$\tan 30^{\circ} = \dfrac {AB}{BD}$

$\dfrac {1}{\sqrt {3}} = \dfrac {AB}{x + 20}$

$x + 20 = \sqrt {3}AB ..... (1)$

In triangle

$ABC$ ,

$\tan 60^{\circ} = \dfrac {AB}{BC}$

$\sqrt {3} = \dfrac {AB}{x}$

$x = \dfrac {AB}{\sqrt {3}} ....(2)$

From (1)

$\dfrac {AB}{\sqrt {3}} + 20 = \sqrt {3} AB$

$AB + 30\sqrt {3} = 3AB$

$2AB = 20\sqrt {3}$

$AB = \dfrac {20\sqrt {3}}{2}$

$= 10\sqrt {3} = 17.32\ cm$

From (2)

$x = \dfrac {AB}{\sqrt {3}} = \dfrac {17.32}{\sqrt {3}} = 10\ cm$ .

Find $PQ$ , if $AB = 150\ m, \angle P = 30^{\circ}$ and $\angle Q = 45^{\circ}$ .

From triangle

$APB$

$\tan 30^{\circ} = \dfrac {AB}{PB}$

$\dfrac {1}{\sqrt {3}} = \dfrac {150}{PB}$

$PB = 150\sqrt {3}$

$= 259.80\ m$

From triangle

$ABQ$

$\tan 45^{\circ} = \dfrac {AB}{BQ}$

$1 = \dfrac {150}{BQ}$

$BQ = 150\ m$

$PQ = PB - BQ$

$= 259.80 - 150$

$= 109.80\ m$ .

Refer given figure, Find $AB$ and $BC$ .

Let

$BC = xm$

$BD = BC + CD = (x + 20) cm$

In triangle

$ABD$ ,

$\tan 45^{\circ} = \dfrac {AB}{BD}$

$1 = \dfrac {AB}{x + 20}$

$x + 20 = AB ..... (1)$

In triangle

$ABC$ ,

$\tan 60^{\circ} = \dfrac {AB}{BC}$

$\sqrt {3} = \dfrac {AB}{x}$

$x = \dfrac {AB}{\sqrt {3}} ..... (2)$

From (1)

$\dfrac {AB}{\sqrt {3}} + 20 = AB$

$AB + 20\sqrt {2} = \sqrt {3} AB$

$AB (\sqrt {3} - 1) = 20\sqrt {3}$

$AB = \dfrac {20\sqrt {2}}{(\sqrt {3} - 1)}$

$= \dfrac {20\sqrt {3}}{(\sqrt {3} - 1)}\times \dfrac {(\sqrt {3} + 1)}{(\sqrt {3} + 1)}$

$= \dfrac {20\sqrt {3}(\sqrt {3} + 1)}{3 - 1} = 47.32\ cm$

From (2)

$x = \dfrac {AB}{\sqrt {3}} = \dfrac {47.32}{\sqrt {3}} = 27.32\ cm$

$\therefore BC = x = 27.32\ cm$

$AB = 47.32\ cm, BC = 27.32\ cm$ .

Find $PQ$ , if $AB = 150\ m, \angle P = 30^{\circ}$ and $\angle Q = 45^{\circ}$ .

From triangle

$APB$

$\tan 30^{\circ} = \dfrac {AB}{PB}$

$\dfrac {1}{\sqrt {3}} = \dfrac {150}{PB}$

$PB = 150\sqrt {3} = 259.80\ m$

From triangle

$ABQ$

$\tan 45^{\circ} = \dfrac {AB}{BQ}$

$1 = \dfrac {150}{BQ}$

$BQ = 150\ m$

$PQ = PB + BQ$

$= 259.80 + 150$

$= 409.80\ m$ .

In the given figure, $AB$ and $EC$ are parallel to each other. Sides $AD$ and $BC$ are $2\ cm$ each and are perpendicular to $AB$ .
Given that $\angle AED = 60^{\circ}$ and $\angle ACD = 45^{\circ}$ . Calculate: $AE$ .

From the right triangle

$ADE$

$\sin 60^{\circ} = \dfrac {AD}{AE}$

$\dfrac {\sqrt {3}}{2} = \dfrac {2}{AE}$

$AE = \dfrac {4}{\sqrt {3}}$ .

In the given figure, $AB$ and $EC$ are parallel to each other. Sides $AD$ and $BC$ are $2\ cm$ each and are perpendicular to $AB$ .
Given that $\angle AED = 60^{\circ}$ and $\angle ACD = 45^{\circ}$ . Calculate $AC$ .

From given figure,

$\sin 45^{\circ} = \dfrac {AD}{AC}$

$\dfrac {1}{\sqrt {2}} = \dfrac {2}{AC}$

$AC = 2\sqrt {2}$ .

Find $AC$ .(Refer figure)

From right angled triangle

$ABC$

$\sin B = \dfrac {AB}{AC}$

$\sin 30^{\circ} = \dfrac {AB}{AC}$

$\dfrac {1}{2} = \dfrac {12}{AC}$

$AC = 24\ cm$ .

Evaluate:
$\dfrac{\cos55^{0}}{\sin35^{0}}+\dfrac{\cot35^{0}}{\tan55^{0}}$

$\dfrac{\cos55^{0}}{\sin35^{0}}+\dfrac{\cot35^{0}}{\tan55^{0}}$

$=\dfrac{\cos(90^{0}-35^{0})}{\sin35^{0}}+\dfrac{\cot(90^{0}-55^{0})}{\tan55^{0}}$

$=\dfrac{\sin35^{0}}{\sin35^{0}}+\dfrac{\tan55^{0}}{\tan55^{0}}$

$=1+1$

$=2$ .

Evaluate:
$\dfrac{\cot54^{0}}{\tan36^{0}}+\dfrac{\tan20^{0}}{\cot70^{0}}-2$

$\dfrac{\cot54^{0}}{\tan36^{0}}+\dfrac{\tan20^{0}}{\cot70^{0}}-2\\$

$=\dfrac{\cot(90^{0}-36^{0})}{\tan36^{0}}+\dfrac{\tan(90^{0}-70^{0})}{\cot70^{0}}-2\\$

$=\dfrac{\tan36^{0}}{\tan36^{0}}+\dfrac{\cot70^{0}}{\cot70^{0}}-2\\$

$=1+1-2\\$

$=2-2\\$

$=0$ .

Evaluate:
$\sin42^{0}\sin48^{0}-\cos42^{0}\cos48^{0}$

$\sin42^{0}\sin48^{0}-\cos42^{0}\cos48^{0}$

$=\sin(90^{0}-48^{0})\sin48^{0}-\cos(90^{0}-48^{0})\cos48^{0}$

$=\cos48^{0}\sin48^{0}-\sin48^{0}\cos48^{0}$

$=\cos48^{0}\sin48^{0}-\cos48^{0}\sin48^{0}$

$=0$ .

Show that:
$\sin42^{0}\sec48^{0}+\cos42^{0}cosec48^{0}=2$

$L.H.S.$

$=\sin42^{0}\sec48^{0}+\cos42^{0}cosec48^{0}$

$=\sin(90^{0}-48^{0})\times \dfrac{1}{\cos48^{0}}+\cos(90^{0}-18^{0})\times \dfrac{1}{\sin48^{0}}$

$=\cos48^{0}\times \dfrac{1}{\cos48^{0}}+\sin48^{0}\times \dfrac{1}{\sin48^{0}}$

$=1+1$

$=2$

$=R.H.S.$

Evaluate:
$\dfrac{2\tan53^{0}}{\cot37^{0}}-\dfrac{\cot80^{0}}{\tan10^{0}}$

$\dfrac{2\tan53^{0}}{\cot37^{0}}-\dfrac{\cot80^{0}}{\tan10^{0}}\\$

$=\dfrac{2\tan(90^{0}-37^{0})}{\cot37^{0}}-\dfrac{\cot(90^{0}-10^{0})}{\tan10^{0}}\\$

$=\dfrac{2\cot37^{0}}{\cot37^{0}}-\dfrac{\tan10^{0}}{\tan10^{0}}\\$

$=2-1\\$

$=1$ .

Evaluate: $\dfrac{\sec75^{0}}{cosec15^{0}}$

$\dfrac{\sec75^{0}}{cosec15^{0}}\\=\dfrac{\sec(90^{0}-15^{0})}{cosec15^{0}}\\=\dfrac{cosec15^{0}}{cosec15^{0}}\\=1$

Evaluate:
$3\dfrac{\sin72^{0}}{\cos18^{0}}-\dfrac{\sec32^{0}}{cosec58^{0}}$

$3\dfrac{\sin72^{0}}{\cos18^{0}}-\dfrac{\sec32^{0}}{cosec58^{0}}$

$=3\dfrac{\sin(90^{0}-18^{0})}{\cos18^{0}}-\dfrac{\sec(90^{0}-58^{0})}{cosec58^{0}}$

$=3\dfrac{\cos18^{0}}{\cos18^{0}}-\dfrac{cosec58^{0}}{cosec58^{0}}\\=3-1\\=2$ .

Show that:
$\tan10^{0}\tan15^{0}\tan75^{0}\tan80^{0}=1$

$L.H.S.$

$=\tan10^{0}\tan15^{0}\tan75^{0}\tan80^{0}$

$=\tan(90^{0}-80^{0})\tan(90^{0}-75^{0})\tan75^{0}\tan80^{0}$

$=\cot80^{0}\cot75^{0}\tan75^{0}\tan80^{0}$

$=(\cot80^{0}\tan80^{0})(\cot75^{0}\tan75^{0})$

$=(1)(1)$

$=1$

$=R.H.S.$

Evaluate:
$\dfrac{\sin80^{0}}{\cos10^{0}}+\sin59^{0}\sec31^{0}$

$\dfrac{\sin80^{0}}{\cos10^{0}}+\sin59^{0}\sec31^{0}\\$

$=\dfrac{\sin(90^{0}-10^{0})}{\cos10^{0}}+\sin(90^{0}-31^{0})\sec31^{0}\\$

$=\dfrac{\cos10^{0}}{\cos10^{0}}+\dfrac{\cos31^{0}}{\cos31^{0}}\\$

$=1+1\\=2$ .

Evaluate: $\dfrac{\tan47^{0}}{\cot43^{0}}$

$\dfrac{\tan47^{0}}{\cot43^{0}}\\=\dfrac{\tan(90^{0}-43^{0})}{\cot43^{0}}\\=\dfrac{\cot43^{0}}{\cot43^{0}}\\=1$

What is the meaning of trigonometry?

The literal meaning of the word trigonometry is the science of measuring the sides and the angles of triangles.

Evaluate:
$2\dfrac{\tan57^{0}}{\cot33^{0}}-\dfrac{\cot70^{0}}{\tan20^{0}}-\sqrt{2}\cos45^{0}$

$2\dfrac{\tan57^{0}}{\cot33^{0}}-\dfrac{\cot70^{0}}{\tan20^{0}}-\sqrt{2}\cos45^{0}\\$

$=2\dfrac{\tan(90^{0}-33^{0})}{\cot33^{0}}-\dfrac{\cot(90^{0}-20^{0})}{\tan20^{0}}-\sqrt{2}\left ( \dfrac{1}{\sqrt{2}} \right )\\$

$=2\dfrac{\cot33^{0}}{\cot33^{0}}-\dfrac{\tan20^{0}}{\tan20^{0}}-1\\$

$=2-1-1\\$

$=0$ .

Find the value of the following :
$\dfrac{\cos 37^o}{\sin 53^o}$

$\dfrac{\cos 37^o}{\sin 53^o}$

$=\dfrac{\cos (90-53^o)}{\sin 53^o}$

$[\because \cos (90^o- \theta) = s\in \theta]$

$=\dfrac{\cos 53^o}{\sin 53^o}$

$=1$

Evaluate :
$cos 48^0 - sin 42^0$

$cos 48^0 - sin 42^0$

$= cos ( 90 -42) - sin 42^0$

$= sin 42^0 - sin 42^0$

$= 0$

Find the value of the following :
$\dfrac{\cos 19^o}{\sin 71^o}$

$=\dfrac{\cos (90^o- 71^o)}{\sin 71^o}$

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

$=\dfrac{\sin 71^o}{\sin 71^o}$

$=1$

Find the value of the following :
$\cot 34^o- \tan 56^o$

$\cot (90^o-56^o)-tan 56^o [\because \cot (90^o - \theta)= \tan \theta]$

$\tan 56^o - \tan 56^o$

$=0$

Evaluate:
$\dfrac{\cos70^{0}}{\sin20^{0}}+\dfrac{\cos59^{0}}{\sin31^{0}}-8\sin^{2}30^{0}$

1888239_1842746_ans_c12441aff5be4ce292c0a4b6f5f4d438.jpeg

Find the value of the following :
$\dfrac{\tan 10^o}{\cot 80^o}$

$\dfrac{\tan 10^o}{\cot 80^o}$

$=\dfrac{\tan (90^o-80^o)}{\cot 80^o}$

$[\because \tan (90^o- \theta) = \cot \theta]$

$=\dfrac{\cot 80^o}{\cot 80^o}$

$=1$

Evaluate:
$\left( \dfrac{\sin 27^o}{\cos 63^o} \right)^2 + \left( \dfrac{\cos 63^o}{\sin 27^o}\right)^2$

$\left( \dfrac{\sin 27^o}{\cos 63^o} \right)^2 + \left( \dfrac{\cos 63^o}{\sin 37^o}\right)^2$

$\left[ \dfrac{\sin 27^o}{\cos (90^o-27^o)} \right]^2 + \left[ \dfrac{\cos 63^o}{\sin (90^o-63^o)}\right]^2$

$\left( \dfrac{\sin 27^o}{\sin 27^o} \right)^2 + \left( \dfrac{\cos 63^o}{\cos 63^o}\right)^2$

$=1+1=2$

Prove that
$\sin \theta \, cosec \theta + \cos \theta \sec \theta = 2$ .

L.H.S

$= \sin \theta . cosec \theta + \cos \theta$ .

$= \sin \theta . cosec \theta + \cos \theta . \sec \theta$

$= \sin \theta . \dfrac{1}{\sin \theta} + \cos \theta \cdot \dfrac{1}{\cos \theta}$

$= 1 + 1$

$= 2$ = R.H.S
Hence Proved.

Solve:
$\sqrt {\dfrac {\sec \theta +1}{\sec \theta -1}}=\cot \theta +\csc \theta$

$L.H.S=\sqrt {\dfrac {\sec \theta +1}{\sec \theta -1}}$

$=\sqrt {\dfrac {\dfrac {1}{\cos \theta}+1}{\dfrac {1}{\cos \theta }-1}}=\sqrt {\dfrac {1+\cos \theta }{1-\cos \theta }}$
Multiply numerator and denominator by

$(1+\cos \theta )$

$=\sqrt {\dfrac {(1+\cos \theta )(1+\cos \theta )}{(1-\cos \theta )(1+\cos \theta )}}$

$=\sqrt {\dfrac {(1+\cos \theta )^2}{1-\cos^2 \theta }}$

$=\sqrt {\dfrac {(1+\cos \theta )^2}{\sin^2 \theta }}$

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

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

$\csc \theta +\cot \theta$ or

$\cot \theta +\csc \theta$

$R.H.S$

Find the value of the following :
$\dfrac{\sin 36^o}{\cos 54^o} - \dfrac{\sin 54^o}{\cos 36^o}$

$=\dfrac{\sin (90^o-54^o)}{\cos 54^o} - \dfrac{\sin (90^o-36^o)}{\cos 36^o}$

$=\dfrac{\cos 54^o}{\cos 54^o}-\dfrac{\cos 36^o}{\cos 36^o}$

$=1-1$

$=0$

Evaluate:
$\sin 70^o \sin 20^o - \cos 20^o \cos 70^o$

$\sin 70^o \sin 20^o - \cos 20^o \cos 70^o$

$\sin (90^o-20^o) \sin 20^o - \cos 20^o \cos (90^o-20^o)$

$=\cos 20^o \sin 20^o- \cos 20^o \sin 20^o$

$=0$

Evaluate:
$\dfrac{2 \cos 67^o}{\sin 23^o}- \dfrac{\tan 40^o}{\cot 50^o}- \cos 60^o$

$\dfrac{2 \cos 67^o}{\sin 23^o}- \dfrac{\tan 40^o}{\cot 50^o}- \cos 60^o$

$\dfrac{2 \cos (90^o-23^o)}{\sin 23^o}- \dfrac{\tan (90^o-50^o)}{\cot 50^o}- \cos 60^o -\dfrac{1}{2}$

$=\dfrac{2 \sin 23^o}{\sin 23^o}- \dfrac{\cot 50^o}{\cot 50^o} -\dfrac{1}{2}$

$=2-1-\dfrac{1}{2}$

$\left[ \because \cos 60^o =\dfrac{1}{2} \right]$

$=2-3/2$

$=\dfrac{1}{2}$

Evaluate:
$\left( \dfrac{\sin 35^o}{\cos 55^o} \right)^2 + \left( \dfrac{\cos 55^o}{\sin 35^o}\right)^2 -2 \cos 60^o$

$\left( \dfrac{\sin 35^o}{\cos 55^o} \right)^2 + \left( \dfrac{\cos 55^o}{\sin 35^o}\right)^2 -2 \cos 60^o$

$\left[ \dfrac{\sin 35^o}{\cos (90^o-35^o)} \right]^2 + \left[ \dfrac{\cos 55^o}{\sin (90^o-55^o)}\right]^2 -2 \cos 60^o$

$\left( \dfrac{\sin 35^o}{\cos 35^o} \right)^2 + \left( \dfrac{\cos 55^o}{\sin 55^o}\right)^2 -2 \times \dfrac{1}{2}$

$=1+1-1=1$

Prove :
$\tan \theta + \tan \left( {{{90}^ \circ } - \theta } \right) = \sec \theta \sec \left( {{{90}^ \circ } - \theta } \right)$

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

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

$=\cfrac{1}{\sin\theta \cos\theta}$

$=\csc\theta\sec\theta$

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

If $\displaystyle \frac {9x}{\cos\theta}+\frac {5y}{\sin\theta}=56$ and $\displaystyle \frac {9x \sin\theta}{\cos^2\theta}-\cfrac {5y \cos\theta}{\sin^2\theta}=0$ , then the value of $[(9x)^{2/3}+(5y)^{2/3}]^3$ is

Given,

$\displaystyle \cfrac { 9x }{ \cos\theta } +\cfrac { 5y }{ \sin \theta } =56$ .....(1)

Also given,

$\displaystyle \cfrac { 9x\sin \theta }{ \cos^{ 2 }\theta } -\cfrac { 5y\cos\theta }{ \sin ^{ 2 }\theta } =0$

$9x \sin ^3\theta=5y \cos^3\theta$ .

$\Rightarrow \displaystyle \cfrac {\cos^3\theta}{9x}=\cfrac {\sin ^3\theta}{5y}=k^3$ (say)

$\Rightarrow \displaystyle \cos\theta=k(9x)^{1/3}$ and

$\\sin \theta=k(5y)^{1/3}$ ....(2)

Squaring and adding, we get

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

$\Rightarrow k^2[(9x)^{2/3}+(5y)^{2/3}]=1$

$\Rightarrow \displaystyle k^2=\cfrac{1}{[(9x)^{2/3}+(5y)^{2/3}]}$

Substitute the value from (2) in (1), we get

$\displaystyle \cfrac {9x}{k(9x)^{1/3}}+\cfrac {5y}{k(5y)^{1/3}}=56$ (by (1))

$\Rightarrow (9x)^{2/3}+(5y)^{2/3}=56k$

$\Rightarrow [(9x)^{2/3}+(5y)^{2/3}]^2=(56)^2k^2$

$\Rightarrow \displaystyle [(9x)^{2/3}+(5y)^{2/3}]^2 =\cfrac {(56)^2}{(9x)^{2/3}+(5y)^{2/3}}$

$\Rightarrow [(9x)^{2/3}+(5y)^{2/3}]^3=(56)^2=3136$ .

Express the trigonometric ratios $\sin A, \sec A$ and $\tan A$ in terms of $\cot A$ .

We know,

$\text{cosec}^2A=1+\cot^2A$

$\dfrac{1}{\text{cosec}^2A}=\dfrac{1}{1+\cot^2A}$

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

$\sin A=\pm\dfrac{1}{\sqrt{1+\cot^2A}}$

$\therefore \sin A=\dfrac{1}{\sqrt{1+\cot^2A}}$

We know,

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

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

$=\dfrac{\cot^2A+1}{\cot^2A}$

$\therefore \sec A=\dfrac{\sqrt{\cot^2A+1}}{\cot A}$

We know,

$\tan A = \dfrac {1}{\cot A}$

Introduction To Trigonometry - Class 10 Maths - Extra Questions

Find value of 10x

Value of 100x is 3464,

From the given figure, find the value of m if, tan C is m4\displaystyle \frac{m}{4}.

\displaystyle\,\frac{1\,-\,cos\,2\,A}{sin\,2\,A}\,=\,tan\,2A\displaystyle\,\frac{1\,-\,cos\,2\,A}{sin\,2\,A}\,=\,tan\,2AIf above statement is true then mention answer as 1, else mention 0 if false

Find side cc of the triangle ABCABC, given that \displaystyle A=30^{0},B=60^{0}\displaystyle A=30^{0},B=60^{0} and \displaystyle b=20\sqrt{3}\displaystyle b=20\sqrt{3}

If 2\cos { \theta } =12\cos { \theta } =1 and \theta\theta is acute angle, then what is \theta \theta equal to?

Find the value of the following:\tan ^{ 2 }{ { 60 }^{ o } } +2\tan ^{ 2 }{ { 45 }^{ o } } \tan ^{ 2 }{ { 60 }^{ o } } +2\tan ^{ 2 }{ { 45 }^{ o } }

Prove that \cos { \theta } \cdot cosec \ { \theta } =\cot { \theta } \cos { \theta } \cdot cosec \ { \theta } =\cot { \theta }

Find the value of \sin ^{ 2 }{ { 45 }^{ o } } +\cos ^{ 2 }{ { 45 }^{ o } } \sin ^{ 2 }{ { 45 }^{ o } } +\cos ^{ 2 }{ { 45 }^{ o } } .

Find the value of { \left( \sin { \theta } +\cos { \theta } \right) }^{ 2 }+{ \left( \sin { \theta } -\cos { \theta } \right) }^{ 2 }{ \left( \sin { \theta } +\cos { \theta } \right) }^{ 2 }+{ \left( \sin { \theta } -\cos { \theta } \right) }^{ 2 }

If \cos{\theta}=1\cos{\theta}=1, then find the value of acute angle \theta\theta.

\sin{\left({60}^{\circ}+{30}^{\circ}\right)}=\sin{\left({60}^{\circ}+{30}^{\circ}\right)}= _______.

(1 + \tan A \cdot \tan B)^{2} + (\tan A - \tan B)^{2} = \sec^{2}A \cdot \sec^{2}B(1 + \tan A \cdot \tan B)^{2} + (\tan A - \tan B)^{2} = \sec^{2}A \cdot \sec^{2}B.

Find the value of \sin^{2}30^{\circ} + \cos^{2} 60^{\circ}\sin^{2}30^{\circ} + \cos^{2} 60^{\circ}

Prove that \sin^2A + \dfrac{1}{1 + \tan^2A} = 1\sin^2A + \dfrac{1}{1 + \tan^2A} = 1.

Prove that (1 + \cot^2A) \sin^2A = 1(1 + \cot^2A) \sin^2A = 1.

Prove the following :(iv) \dfrac{\cos(90^{\circ} \, - \, A) \sin (90^{\circ} \, - \, A)}{\tan(90^{\circ} \, - \, A)} = \, \sin^2 \, A\dfrac{\cos(90^{\circ} \, - \, A) \sin (90^{\circ} \, - \, A)}{\tan(90^{\circ} \, - \, A)} = \, \sin^2 \, A

Prove that cosec \theta \sqrt{1 - \cos^2 \theta} = 1cosec \theta \sqrt{1 - \cos^2 \theta} = 1.

Evaluate the following :\dfrac{cos 37^{\circ}}{sin 53^{\circ}}\dfrac{cos 37^{\circ}}{sin 53^{\circ}}

Evaluate the following :\dfrac{\tan 10}{\cot 80}\dfrac{\tan 10}{\cot 80}

Evaluate the following :\dfrac{\sec 11}{\text{cosec} 79}\dfrac{\sec 11}{\text{cosec} 79}

Prove that (\sec \theta + \cos \theta) (\sec \theta - \cos \theta) = \tan^2 \theta + \sin^2\theta(\sec \theta + \cos \theta) (\sec \theta - \cos \theta) = \tan^2 \theta + \sin^2\theta.

Prove that (\sec^2\theta - 1) (co\sec^2\theta - 1) = 1(\sec^2\theta - 1) (co\sec^2\theta - 1) = 1

Evaluate the following :\dfrac{tan 54^{\circ}}{cot 36^{\circ}}\dfrac{tan 54^{\circ}}{cot 36^{\circ}}

Evaluate the following:\dfrac{2 \tan 53^{\circ}}{\cot 37^{\circ}} \, - \, \dfrac{\cot 80^{\circ}}{\tan 10^{\circ}}\dfrac{2 \tan 53^{\circ}}{\cot 37^{\circ}} \, - \, \dfrac{\cot 80^{\circ}}{\tan 10^{\circ}}

Verify that \cot ^{ 2 }{ {60}^{o} } +\sin ^{ 2 }{ {45}^{o} } +\sin ^{ 2 }{ {30}^{o} } +\cos ^{ 2 }{ {90}^{o} } =\dfrac { 13 }{ 12 }\cot ^{ 2 }{ {60}^{o} } +\sin ^{ 2 }{ {45}^{o} } +\sin ^{ 2 }{ {30}^{o} } +\cos ^{ 2 }{ {90}^{o} } =\dfrac { 13 }{ 12 }

If \tan { \theta } =\dfrac { 20 }{ 21 }\tan { \theta } =\dfrac { 20 }{ 21 }, then prove that \dfrac { 1-\sin { \theta } +\cos { \theta } }{ 1+\sin { \theta } +\cos { \theta } } =\dfrac { 3 }{ 7 }\dfrac { 1-\sin { \theta } +\cos { \theta } }{ 1+\sin { \theta } +\cos { \theta } } =\dfrac { 3 }{ 7 }

Express each of the following in trigonometric ratios of angles between 0^{\circ} \, and \, 45^{\circ}0^{\circ} \, and \, 45^{\circ} :sin 85^{\circ} \, + \, cosec 85^{\circ}sin 85^{\circ} \, + \, cosec 85^{\circ}

{ \left( \tan { x } +\cot { x } \right) }^{ 2 }=\sec ^{ 2 }{ x+\csc ^{ 2 }{ x } }{ \left( \tan { x } +\cot { x } \right) }^{ 2 }=\sec ^{ 2 }{ x+\csc ^{ 2 }{ x } }

Prove that:\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos\,A}\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos\,A}

Calculate : { \left( \cos { 40 } \right) }^{ 2 }+{ \left( \sin { 40 } \right) }^{ 2 }{ \left( \cos { 40 } \right) }^{ 2 }+{ \left( \sin { 40 } \right) }^{ 2 }

Prove that \dfrac{{\cot A - \cos A}}{{\cot A + \cos A}} = \dfrac{{\cos ecA - 1}}{{\cos ecA + 1}}\dfrac{{\cot A - \cos A}}{{\cot A + \cos A}} = \dfrac{{\cos ecA - 1}}{{\cos ecA + 1}}

Prove that: \dfrac{1}{\sec A-1}+\dfrac{1}{\sec A+1}=2 \csc A \cot A\dfrac{1}{\sec A-1}+\dfrac{1}{\sec A+1}=2 \csc A \cot A

Prove that \dfrac {1+\cos \theta}{\sin^{2}\theta}=\dfrac {1}{1-\cos \theta}\dfrac {1+\cos \theta}{\sin^{2}\theta}=\dfrac {1}{1-\cos \theta}

Evaluate\dfrac{tan 45^o}{cosec 30^o} + \dfrac{sec 60^o}{cot 45^o} - \dfrac{5 sin 90^o}{2 cos 0^o}\dfrac{tan 45^o}{cosec 30^o} + \dfrac{sec 60^o}{cot 45^o} - \dfrac{5 sin 90^o}{2 cos 0^o}

Evaluate : \sin^{2}90.\cos^{2} 45+4 \tan^{2}30+\dfrac {1}{2}\sin^{2}a-2 \cos^{2}90+\dfrac {1}{24}\sin^{2}90.\cos^{2} 45+4 \tan^{2}30+\dfrac {1}{2}\sin^{2}a-2 \cos^{2}90+\dfrac {1}{24}

Simplify:\dfrac {\sin \theta.\csc (90-\theta).\tan \theta}{\csc (90-\theta).\cos \theta.\cot (90-\theta)}\dfrac {\sin \theta.\csc (90-\theta).\tan \theta}{\csc (90-\theta).\cos \theta.\cot (90-\theta)}

For any angle of measure \theta\theta prove that 1+\tan^2\theta =\sec^2\theta1+\tan^2\theta =\sec^2\theta.

Find the value of \dfrac{\cos 30^o \tan 60^o + \sin 60^o \cos 60^o}{cosec \,30^o \sec 60^o - cosec \,60^o \cot 30^o}\dfrac{\cos 30^o \tan 60^o + \sin 60^o \cos 60^o}{cosec \,30^o \sec 60^o - cosec \,60^o \cot 30^o}

Prove that: \sqrt {\sec^{2}A+\csc^{2}A}=\tan A+\cot A\sqrt {\sec^{2}A+\csc^{2}A}=\tan A+\cot A

If A = 15^oA = 15^o, verify that 4 \sin 2 A . \cos 4 A . \sin 6 A = 14 \sin 2 A . \cos 4 A . \sin 6 A = 1

Evaluate\sin^{2}\theta(\tan^{2}\theta+1)=\sec^{2}\theta-1\sin^{2}\theta(\tan^{2}\theta+1)=\sec^{2}\theta-1

Evaluate\tan\theta.\sin\theta+\cos\theta=\sec\theta\tan\theta.\sin\theta+\cos\theta=\sec\theta

In the figure, find the value of BCBC

Solve \dfrac{{\cos {{45}^ \circ }}}{{\sec {{30}^ \circ } + {\mathop{\rm cosec}\nolimits} {{30}^ \circ }}}\dfrac{{\cos {{45}^ \circ }}}{{\sec {{30}^ \circ } + {\mathop{\rm cosec}\nolimits} {{30}^ \circ }}}

Evaluate: \dfrac{\cos\ 3A-2\cos(4A)}{\sin\ 3A+2\sin(4A)}\dfrac{\cos\ 3A-2\cos(4A)}{\sin\ 3A+2\sin(4A)}; when A=15^{o}A=15^{o}

Solve:-\dfrac{{2\tan {{45}^ \circ }}}{{1 + {{\tan }^2}{{45}^ \circ }}}\dfrac{{2\tan {{45}^ \circ }}}{{1 + {{\tan }^2}{{45}^ \circ }}}

Prove that \cos \theta \sin(90^o-\theta)+\sin \theta \cos(90^o-\theta)=1\cos \theta \sin(90^o-\theta)+\sin \theta \cos(90^o-\theta)=1

Without using trigonometric tables, evaluate the following: \left ( \sin 35^{0}\cos 55^{0}+\cos 35^{0}\sin 55^{0} \right )/\left ( cosec^{2}10^{0}-\tan ^{2}80^{0} \right )\left ( \sin 35^{0}\cos 55^{0}+\cos 35^{0}\sin 55^{0} \right )/\left ( cosec^{2}10^{0}-\tan ^{2}80^{0} \right ).

Prove that following identities, where the angles involved are acute angles for which the trigonometric ratios as defined: \left ( 1+\tan ^{2}A \right )\left ( 1-\sin A \right )\left ( 1+\sin A \right )=1\left ( 1+\tan ^{2}A \right )\left ( 1-\sin A \right )\left ( 1+\sin A \right )=1.

Verify that\tan^{2} 30^{\circ} + \tan^{2} 45^{\circ} + \tan^{2} 60^{\circ} = 4\dfrac {1}{3}\tan^{2} 30^{\circ} + \tan^{2} 45^{\circ} + \tan^{2} 60^{\circ} = 4\dfrac {1}{3}.

A 15 m15 m long ladder touches the top of vertical wall. If this ladder makes an angle of 60^{0}60^{0} with the wall, then height of the wall is :

Prove that following identities, where the angles involved are acute angles for which the trigonometric ratios as defined: 1/\left ( 1+\cos A \right )+1/\left ( 1-\cos A \right )=2cosec^{2}A1/\left ( 1+\cos A \right )+1/\left ( 1-\cos A \right )=2cosec^{2}A.

Verify that\cos 45^{\circ} \cos 60^{\circ} - \sin 45^{\circ} \sin 60^{\circ} = -\dfrac {\sqrt {3} - 1}{2\sqrt {2}}\cos 45^{\circ} \cos 60^{\circ} - \sin 45^{\circ} \sin 60^{\circ} = -\dfrac {\sqrt {3} - 1}{2\sqrt {2}}.

Verify thatcosec^{2} 45^{\circ} \cdot \sec^{2} 30^{\circ} \cdot \sin^{3} 90^{\circ} \cdot \cos 60^{\circ} = 1\dfrac {1}{3}cosec^{2} 45^{\circ} \cdot \sec^{2} 30^{\circ} \cdot \sin^{3} 90^{\circ} \cdot \cos 60^{\circ} = 1\dfrac {1}{3}.

Find the value of \cfrac{\tan 47^{0}}{\cot 43^{0}}\cfrac{\tan 47^{0}}{\cot 43^{0}}

Simplify: (1 + \tan^{2} \theta) (1 - \sin \theta) (1 + \sin \theta)(1 + \tan^{2} \theta) (1 - \sin \theta) (1 + \sin \theta).

State true or false:\sin^2A+ \cos^2A= 1\sin^2A+ \cos^2A= 1

If \sin A, \cos A\sin A, \cos A and \tan A\tan A are in geometric progression, then \cot^6A-\cot^2A=........\cot^6A-\cot^2A=........

\tan 30^{\circ} \: \cot 30^{\circ}=\tan 30^{\circ} \: \cot 30^{\circ}=

\text{cosec}\, ^{2} 60^{0}-\tan ^{2} 30^{0}=\text{cosec}\, ^{2} 60^{0}-\tan ^{2} 30^{0}=

If the angle \theta = 60^{\circ}\theta = 60^{\circ} and \displaystyle cos \theta = \frac{1}{2}\displaystyle cos \theta = \frac{1}{2}, find the value of sec \theta\theta.

Find the value of : \displaystyle \dfrac{2 \tan 53^o}{\cot 37^o} - \dfrac{\cot 80^o}{\tan 10^o}\displaystyle \dfrac{2 \tan 53^o}{\cot 37^o} - \dfrac{\cot 80^o}{\tan 10^o}

If \displaystyle \sin \theta = \cfrac{\sqrt{2}}{3}, \cos \theta = \cfrac{1}{3}\displaystyle \sin \theta = \cfrac{\sqrt{2}}{3}, \cos \theta = \cfrac{1}{3}, then \displaystyle \tan \theta = \sqrt{m}.\displaystyle \tan \theta = \sqrt{m}. The value of mm is

If \displaystyle tan \theta = \frac{2}{5}\displaystyle tan \theta = \frac{2}{5}, then \displaystyle cot \theta = \frac{p}{2}\displaystyle cot \theta = \frac{p}{2}, pp is

If \tan 2A = \cot (A - 18^o)\tan 2A = \cot (A - 18^o) then find the value of A where (2A) and (A - 18^o)(A - 18^o) are acute angles

The value of \displaystyle \left ( \dfrac{\tan 60^{0}+1}{\tan 60^{0}-1} \right )^{2}= \dfrac{1+\cos 30^{0}}{1-\cos 30^{0}}\displaystyle \left ( \dfrac{\tan 60^{0}+1}{\tan 60^{0}-1} \right )^{2}= \dfrac{1+\cos 30^{0}}{1-\cos 30^{0}}if true then enter 11 and if false then write 00

The value of \displaystyle \tan \left ( 2 \times 30^{0}\right )= \frac{2 \tan 30^{0}}{1-\tan ^{2} 30^{0}}\displaystyle \tan \left ( 2 \times 30^{0}\right )= \frac{2 \tan 30^{0}}{1-\tan ^{2} 30^{0}}if true then enter 11 and if false then write 00

The value of \cos ^{2} 30^{0}-\sin ^{2} 30^{0}= \cos 60^o\cos ^{2} 30^{0}-\sin ^{2} 30^{0}= \cos 60^oIf above statement is true then enter 11 and if false then write 00.

\sin ^{2} 30^{0}+\cos ^{2} 30^{0}\sin ^{2} 30^{0}+\cos ^{2} 30^{0}

The value of \cos 30^{\circ} \cos 60^{\circ}-\sin 30^{\circ} \sin 60^{\circ} \cos 30^{\circ} \cos 60^{\circ}-\sin 30^{\circ} \sin 60^{\circ} is equal to

ABCABC is an isosceles right-angled triangle. Assuming AB= BC= xAB= BC= x, find the value of each of the following trigonometric ratios:\sin 45^{0}\sin 45^{0} is \displaystyle \frac{1}{\sqrt{2}}\displaystyle \frac{1}{\sqrt{2}}If true then enter 11 and if false then enter 00

Find the value of \text{cosec}^{2} 45^{o}-\cot^{2} 45^{o}\text{cosec}^{2} 45^{o}-\cot^{2} 45^{o}.

The value of \displaystyle \cos \left ( 2 \times 30^{0}\right )= \frac{1-\tan ^{2} 30^{0}}{1+\tan ^{2} 30^{0}}\displaystyle \cos \left ( 2 \times 30^{0}\right )= \frac{1-\tan ^{2} 30^{0}}{1+\tan ^{2} 30^{0}}If true then enter 11 and if false then write 00

The value of \displaystyle \sin \left ( 2 \times 30^{0}\right )= \frac{2\tan 30^{0}}{1+\tan ^{2} 30^{0}}\displaystyle \sin \left ( 2 \times 30^{0}\right )= \frac{2\tan 30^{0}}{1+\tan ^{2} 30^{0}}if true then enter 11 and if false then write 00

If A= 30^{0}A= 30^{0}, then the value of\sin \left ( 60^{0}+A \right )-\sin \left ( 60^{0}-A \right )= \sin A\sin \left ( 60^{0}+A \right )-\sin \left ( 60^{0}-A \right )= \sin AIf true then enter 11 and if false then write 00

If \tan \theta= \cot \theta \tan \theta= \cot \theta and 0^{0}\leq \theta \leq 90^{0}0^{0}\leq \theta \leq 90^{0}, state the value of \theta\theta in degrees

The value of \sin 60^{0}=2 \sin 30^{0} \cos 30^{0}\sin 60^{0}=2 \sin 30^{0} \cos 30^{0}if true then enter 11 and if false then write 00

If \sqrt{3}= 1.732\sqrt{3}= 1.732, (correct to two decimal places) the value of \sin 60^{0}\sin 60^{0} is 0.87.0.87.If true then enter 11 and if false then enter 00.

From the given figure, find the value of m if, tan C is $\displaystyle \frac{m}{4}$ .

$\displaystyle\,\frac{1\,-\,cos\,2\,A}{sin\,2\,A}\,=\,tan\,2A$

If above statement is true then mention answer as 1, else mention 0 if false

Find side $c$ of the triangle $ABC$ , given that $\displaystyle A=30^{0},B=60^{0}$ and $\displaystyle b=20\sqrt{3}$

If $2\cos { \theta } =1$ and $\theta$ is acute angle, then what is $\theta$ equal to?

Find the value of the following:
$\tan ^{ 2 }{ { 60 }^{ o } } +2\tan ^{ 2 }{ { 45 }^{ o } }$

Prove that $\cos { \theta } \cdot cosec \ { \theta } =\cot { \theta }$

Find the value of $\sin ^{ 2 }{ { 45 }^{ o } } +\cos ^{ 2 }{ { 45 }^{ o } }$ .

Find the value of ${ \left( \sin { \theta } +\cos { \theta } \right) }^{ 2 }+{ \left( \sin { \theta } -\cos { \theta } \right) }^{ 2 }$

If $\cos{\theta}=1$ , then find the value of acute angle $\theta$ .

$\sin{\left({60}^{\circ}+{30}^{\circ}\right)}=$ _______.

$(1 + \tan A \cdot \tan B)^{2} + (\tan A - \tan B)^{2} = \sec^{2}A \cdot \sec^{2}B$ .

Find the value of $\sin^{2}30^{\circ} + \cos^{2} 60^{\circ}$

Prove that $\sin^2A + \dfrac{1}{1 + \tan^2A} = 1$ .

Prove that $(1 + \cot^2A) \sin^2A = 1$ .

Prove the following :
(iv) $\dfrac{\cos(90^{\circ} \, - \, A) \sin (90^{\circ} \, - \, A)}{\tan(90^{\circ} \, - \, A)} = \, \sin^2 \, A$

Prove that $cosec \theta \sqrt{1 - \cos^2 \theta} = 1$ .

Evaluate the following :
$\dfrac{cos 37^{\circ}}{sin 53^{\circ}}$

Evaluate the following :
$\dfrac{\tan 10}{\cot 80}$

Evaluate the following :

$\dfrac{\sec 11}{\text{cosec} 79}$

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

Prove that $(\sec^2\theta - 1) (co\sec^2\theta - 1) = 1$

Evaluate the following :
$\dfrac{tan 54^{\circ}}{cot 36^{\circ}}$

Evaluate the following:

$\dfrac{2 \tan 53^{\circ}}{\cot 37^{\circ}} \, - \, \dfrac{\cot 80^{\circ}}{\tan 10^{\circ}}$

Verify that $\cot ^{ 2 }{ {60}^{o} } +\sin ^{ 2 }{ {45}^{o} } +\sin ^{ 2 }{ {30}^{o} } +\cos ^{ 2 }{ {90}^{o} } =\dfrac { 13 }{ 12 }$

If $\tan { \theta } =\dfrac { 20 }{ 21 }$ , then prove that $\dfrac { 1-\sin { \theta } +\cos { \theta } }{ 1+\sin { \theta } +\cos { \theta } } =\dfrac { 3 }{ 7 }$

Express each of the following in trigonometric ratios of angles between $0^{\circ} \, and \, 45^{\circ}$ :
$sin 85^{\circ} \, + \, cosec 85^{\circ}$

${ \left( \tan { x } +\cot { x } \right) }^{ 2 }=\sec ^{ 2 }{ x+\csc ^{ 2 }{ x } }$

Prove that: $\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos\,A}$

Calculate : ${ \left( \cos { 40 } \right) }^{ 2 }+{ \left( \sin { 40 } \right) }^{ 2 }$

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

Prove that: $\dfrac{1}{\sec A-1}+\dfrac{1}{\sec A+1}=2 \csc A \cot A$

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

Evaluate
$\dfrac{tan 45^o}{cosec 30^o} + \dfrac{sec 60^o}{cot 45^o} - \dfrac{5 sin 90^o}{2 cos 0^o}$

Evaluate : $\sin^{2}90.\cos^{2} 45+4 \tan^{2}30+\dfrac {1}{2}\sin^{2}a-2 \cos^{2}90+\dfrac {1}{24}$

Simplify:
$\dfrac {\sin \theta.\csc (90-\theta).\tan \theta}{\csc (90-\theta).\cos \theta.\cot (90-\theta)}$

For any angle of measure $\theta$
prove that $1+\tan^2\theta =\sec^2\theta$ .

Find the value of $\dfrac{\cos 30^o \tan 60^o + \sin 60^o \cos 60^o}{cosec \,30^o \sec 60^o - cosec \,60^o \cot 30^o}$

Prove that: $\sqrt {\sec^{2}A+\csc^{2}A}=\tan A+\cot A$

If $A = 15^o$ , verify that $4 \sin 2 A . \cos 4 A . \sin 6 A = 1$

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

Evaluate
$\tan\theta.\sin\theta+\cos\theta=\sec\theta$

In the figure, find the value of $BC$

Solve $\dfrac{{\cos {{45}^ \circ }}}{{\sec {{30}^ \circ } + {\mathop{\rm cosec}\nolimits} {{30}^ \circ }}}$

Evaluate: $\dfrac{\cos\ 3A-2\cos(4A)}{\sin\ 3A+2\sin(4A)}$ ; when $A=15^{o}$

Solve:-
$\dfrac{{2\tan {{45}^ \circ }}}{{1 + {{\tan }^2}{{45}^ \circ }}}$

Prove that $\cos \theta \sin(90^o-\theta)+\sin \theta \cos(90^o-\theta)=1$

Without using trigonometric tables, evaluate the following: $\left ( \sin 35^{0}\cos 55^{0}+\cos 35^{0}\sin 55^{0} \right )/\left ( cosec^{2}10^{0}-\tan ^{2}80^{0} \right )$ .

Prove that following identities, where the angles involved are acute angles for which the trigonometric ratios as defined: $\left ( 1+\tan ^{2}A \right )\left ( 1-\sin A \right )\left ( 1+\sin A \right )=1$ .

Verify that
$\tan^{2} 30^{\circ} + \tan^{2} 45^{\circ} + \tan^{2} 60^{\circ} = 4\dfrac {1}{3}$ .

A $15 m$ long ladder touches the top of vertical wall. If this ladder makes an angle of $60^{0}$ with the wall, then height of the wall is :

Prove that following identities, where the angles involved are acute angles for which the trigonometric ratios as defined: $1/\left ( 1+\cos A \right )+1/\left ( 1-\cos A \right )=2cosec^{2}A$ .

Verify that
$\cos 45^{\circ} \cos 60^{\circ} - \sin 45^{\circ} \sin 60^{\circ} = -\dfrac {\sqrt {3} - 1}{2\sqrt {2}}$ .

Verify that
$cosec^{2} 45^{\circ} \cdot \sec^{2} 30^{\circ} \cdot \sin^{3} 90^{\circ} \cdot \cos 60^{\circ} = 1\dfrac {1}{3}$ .

Find the value of $\cfrac{\tan 47^{0}}{\cot 43^{0}}$

Simplify: $(1 + \tan^{2} \theta) (1 - \sin \theta) (1 + \sin \theta)$ .

State true or false:
$\sin^2A+ \cos^2A= 1$

If $\sin A, \cos A$ and $\tan A$ are in geometric progression, then $\cot^6A-\cot^2A=........$

$\tan 30^{\circ} \: \cot 30^{\circ}=$

$\text{cosec}\, ^{2} 60^{0}-\tan ^{2} 30^{0}=$

If the angle $\theta = 60^{\circ}$ and find the value of sec $\theta$ .

$\displaystyle \dfrac{2 \tan 53^o}{\cot 37^o} - \dfrac{\cot 80^o}{\tan 10^o}$

If $\displaystyle \sin \theta = \cfrac{\sqrt{2}}{3}, \cos \theta = \cfrac{1}{3}$ , then $\displaystyle \tan \theta = \sqrt{m}.$ The value of $m$ is

If $\displaystyle tan \theta = \frac{2}{5}$ , then $\displaystyle cot \theta = \frac{p}{2}$ , $p$ is

If $\tan 2A = \cot (A - 18^o)$ then find the value of A where (2A) and $(A - 18^o)$ are acute angles

The value of $\displaystyle \left ( \dfrac{\tan 60^{0}+1}{\tan 60^{0}-1} \right )^{2}= \dfrac{1+\cos 30^{0}}{1-\cos 30^{0}}$

if true then enter $1$ and if false then write $0$

The value of $\displaystyle \tan \left ( 2 \times 30^{0}\right )= \frac{2 \tan 30^{0}}{1-\tan ^{2} 30^{0}}$

if true then enter $1$ and if false then write $0$

The value of $\cos ^{2} 30^{0}-\sin ^{2} 30^{0}= \cos 60^o$
If above statement is true then enter $1$ and if false then write $0$ .

$\sin ^{2} 30^{0}+\cos ^{2} 30^{0}$

The value of $\cos 30^{\circ} \cos 60^{\circ}-\sin 30^{\circ} \sin 60^{\circ}$ is equal to

$ABC$ is an isosceles right-angled triangle. Assuming $AB= BC= x$ , find the value of each of the following trigonometric ratios:
$\sin 45^{0}$ is $\displaystyle \frac{1}{\sqrt{2}}$
If true then enter $1$ and if false then enter $0$

Find the value of $\text{cosec}^{2} 45^{o}-\cot^{2} 45^{o}$ .

The value of $\displaystyle \cos \left ( 2 \times 30^{0}\right )= \frac{1-\tan ^{2} 30^{0}}{1+\tan ^{2} 30^{0}}$

If true then enter $1$ and if false then write $0$

The value of $\displaystyle \sin \left ( 2 \times 30^{0}\right )= \frac{2\tan 30^{0}}{1+\tan ^{2} 30^{0}}$

if true then enter $1$ and if false then write $0$

If $A= 30^{0}$ , then the value of
$\sin \left ( 60^{0}+A \right )-\sin \left ( 60^{0}-A \right )= \sin A$
If true then enter $1$ and if false then write $0$

If $\tan \theta= \cot \theta$ and $0^{0}\leq \theta \leq 90^{0}$ , state the value of $\theta$ in degrees

The value of $\sin 60^{0}=2 \sin 30^{0} \cos 30^{0}$

if true then enter $1$ and if false then write $0$

If $\sqrt{3}= 1.732$ , (correct to two decimal places) the value of $\sin 60^{0}$ is $0.87.$
If true then enter $1$ and if false then enter $0$ .

If $x= 15^{0}$ , evaluate:
$\displaystyle 12 \sin 3x \cos 3x+5 \tan ( 2x+ 15^{0})-3 \cot ^{2}3x$

If $\sin x= \cos x$ and $x$ is acute, state the value of $x$ in degrees.

If $A= 60^{0}$ and $B= 30^{0}$ , find the value of $\left ( \sin A \cos B+ \cos A\sin B \right )^{2}+\left ( \cos A \cos B-\sin A \sin B \right )^{2}$