If $$4\ \sin^{2}x-1=0$$ and $$0 < x < 2 \pi$$, then positive values of $$x$$ are

$$30^{o},150^{o},210^{o},330^{o}$$

$$30^{o},120^{o},210^{o},300^{o}$$

$$30^{o},120^{o},150^{o},210^{o}$$

$$30^{o},160^{o},210^{o},320^{o}$$

MCQ Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 10

$\Delta ABC$ , a

$sin(B-C)+b{\,}sin (C-A)+c{\,}sin(A-B)=$

0%
$0$
0%
$a+b+c$
0%
$a^2+b^2+c^2$
0%
$2(a^2+b^2+c^2)$

Explanation

$\triangle ABC,a=2 R\sin A=2 R\sin (180^{\circ}-(B+C))=2 R\sin (B+C)$

$a\sin(B-C)+b\sin (C-A)+c\sin(A-B)=\displaystyle\sum a\sin (B-C)$

$=2 R\sum \sin (B+C)\sin (B-C)$

$=2 R\sum (\sin ^2 B-\sin^2 C)$

$(\because \sin(A+B)\sin (A-B)=\sin^2 A-\sin^2 B)$

$=2 R(\sin^2 B-\sin^2 C+\sin^2 C-\sin^2 A+\sin^2 A-\sin^2 B)$

$=0$

$\sin^2 20+\sin^270$ is equal to_____

0%
$1$
0%
$-1$
0%
$0$
0%
$2$

Explanation

$\sin^2 20^o+\sin^270^o=\sin^2 20^o+\sin^2(90^o-20^o)=\sin^220^o+\cos^220^o=1$

Let

$x$ and

$y$ be

$2$ real numbers which satisfy the equations

$(\tan^{2} x - \sec^{2}y) = \dfrac {5a}{6} - 3$ and

$(-\sec^{2}x + \tan^{2}y) = a^{2}$ , then the value of a can be equal to

0%
$\dfrac {2}{3}$
0%
$\dfrac {-2}{3}$
0%
$\dfrac {3}{2}$
0%
$\dfrac {-3}{2}$

$\sin (A+B) . \sin (A-B) =$

0%
$\sin^2 A - \cos^2 B$
0%
$\cos^2 A - \sin^2 B$
0%
$\sin^2 A - \sin^2 B$
0%
$\cos^2 A - \cos^2 B$

Explanation

$\begin{array}{l} \sin \left( { A+B } \right) .\sin \left( { A-B } \right) \\ =\left[ { \sin \left( A \right) \cos \left( B \right) +\cos \left( A \right) \sin \left( B \right) } \right] \left[ { \sin \left( A \right) \cos \left( B \right) -\cos \left( A \right) \sin \left( B \right) } \right] \\ ={ \sin ^{ 2 } }A{ \cos ^{ 2 } }A-{ \cos ^{ 2 } }A{ \sin ^{ 2 } }A \\ ={ \sin ^{ 2 } }A\left[ { 1-{ { \sin }^{ 2 } }B } \right] -\left[ { 1-{ { \sin }^{ 2 } }A } \right] { \sin ^{ 2 } }B \\ ={ \sin ^{ 2 } }A-{ \sin ^{ 2 } }A{ \sin ^{ 2 } }B-{ \sin ^{ 2 } }B+{ \sin ^{ 2 } }A{ \sin ^{ 2 } }B \\ ={ \sin ^{ 2 } }A-{ \sin ^{ 2 } }B \\ Hence,\, the\, option\, C\, is\, the\, correct\, answer. \end{array}$

$4\ \sin^{2}x-1=0$ and

$0 < x < 2 \pi$ , then positive values of

$x$ are

0%
$30^{o},120^{o},210^{o},300^{o}$
0%
$30^{o},150^{o},210^{o},330^{o}$
0%
$30^{o},120^{o},150^{o},210^{o}$
0%
$30^{o},160^{o},210^{o},320^{o}$

Solution set of the equation

$\sin ^{ 2 }{ x } +\cos ^{ 2 }{ 3x }$ =1$$ is given by

0%
$\left\{ \frac { n\pi }{ 4 } ,n\epsilon I \right\}$
0%
$\left\{ n\pi ,n\epsilon I \right\}$
0%
$\left\{ \frac { n\pi }{ 2 } ,n\epsilon I \right\}$
0%
$none of these$

Explanation

According to question,

$\sin^2x+\cos^23x=1$

$\implies cos^23x= 1-\sin^2x$

$\implies \cos^23x=\cos^2x$

Solving through trigonometric eq,

$3x=2n\pi+x$

$\implies 2x=2n\pi$

$x=n\pi$ ,

$n\epsilon I$

The no of solution of the equation:

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

$x$ ,

$\left[ 0,40 \right]$ are

0%
$4$
0%
$5$
0%
$7$
0%
$None\ of\ these$

Explanation

$1+sinx cosx = 2sinx cosx$

$x\epsilon [ 0,40]$

$1 = \dfrac{2}{2}sinx cosx$

$[\because 2 sinx cosx - sin2x]$

$2 = sin 2x.$

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

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

1168910_1139963_ans_582f026fe627480b9bf890e5f2f03b83.jpeg

$\sin { x } sin{ \left( { 60 }^{ O }+x \right) .\left( { 60 }^{ O }-x \right) }=\frac { 1 }{ 8 }$ then

$x=$

0%
$\pi+(-1)^\alpha\frac{\pi}{6}$
0%
$\frac{\pi}{3}+(-1)^\alpha\frac{\pi}{18}$
0%
$n\pi \left( -1 \right) ^{ \alpha }\frac { \pi }{ 3 }$
0%
$\frac { n\pi }{ 3 } +\left( -1 \right) ^{ \alpha }\frac { \pi }{ 9 }$

Explanation

Given

$\sin x\sin (60^{\circ} +x) \sin (60^{\circ}-x)=\dfrac 18$

$\sin x\left( \sin ^2 60^{\circ} -\sin ^2 x\right)$

$\bigg( \sin (A+B)\sin (A-B) =\sin ^2 A-\sin ^2B\bigg)$

$\sin x\left(\dfrac 34 -\sin ^2 x\right)=\dfrac 18$

$\left(\dfrac {3\sin x-4\sin ^3 x}{4}\right)=\dfrac 18$

$\sin 3x =\dfrac 12$

$\bigg(\sin 3x=3\sin x-4\sin ^3 x\bigg)$

$\implies 3x=n\pi + (-1)^n \dfrac \pi 6$

$\implies x=\dfrac {n \pi}3 + (-1)^n \dfrac \pi {18}$

Find number of solutions to the equation:

$\left[ \sin { x+\cos { x } } \right] =3+\left[ -\sin { x } \right] +\left[ -\cos { x } \right]$

0%
$0$
0%
$1$
0%
$2$
0%
infinite

The number of solutions of the equation

$|\cot x|=\cot x+\dfrac{1}{\sin x}(0\le x\le 2\pi)$ is :

0%
$0$
0%
$1$
0%
$2$
0%
$3$

Explanation

$(\cot x)=\cot x+\dfrac{1}{\sin x}$

$0\le x\le 2\pi$

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

$0\le x\le \dfrac{\pi}{2}$

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

$\sin x=0$

$x=0$

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

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

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

$0\cos x=\dfrac{1}{2}$

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

$\cos 3x=-1$ , where

$0^{o} \ge x \ge 360^{o}$ , then

$x$

0%
$60^{o},180^{o},300^{o}$
0%
$180^{o}$
0%
$60^{o},180^{o}$
0%
$180^{o},300^{o}$

Explanation

Given

$\cos 3 x=-1$

$\implies 3 x=2 n \pi\pm \pi=(2 n \pm 1)\pi$ where

$n$ is an integer

$\implies x=\dfrac{(2 n\pm 1)\pi}{3}$

$\implies x=(2 n\pm 1)60^{\circ}=\pm 60^{\circ},\pm 180^{\circ},\pm 300^{\circ},\pm 420^{\circ},\cdots$

$0^{\circ}\le x\le 360^{\circ},x=60^{\circ},180^{\circ},300^{\circ}$

$\theta$ is an acute angle such that

$\sec^2 \theta = 3$ . then the value of

$\dfrac{\tan^2 \theta - \text{cosec}^2 \theta}{\tan^2 \theta - \text{cosec}^2 \theta}$ is

0%
$\dfrac{4}{7}$
0%
$\dfrac{3}{7}$
0%
$1$
0%
$\dfrac{1}{7}$

Explanation

Given

$\text{sec}^2 \theta=3$

for any value of

$\theta$ ,

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

$(\because \text{the numerator and denominator is same})$

$3\sec\theta+2\sec\dfrac{\pi}{4}=2\cos\theta$ , where

$\theta\in (0, 2\pi)$ , then which of the following can be correct?

0%
$\cos 2\theta=0$
0%
$\sin 2\theta=1$
0%
$\tan \theta=1$
0%
$\tan \theta=-1$

Explanation

$3\sec \theta +2\sec \frac{\pi }{4}=2\cos \theta$

$\Rightarrow \frac{3}{\cos \theta }+2\sqrt{2}=2\cos \theta$

$\Rightarrow \frac{3+2\sqrt{2}\cos \theta }{\cos \theta }=2\cos \theta$

$\Rightarrow 3+2\sqrt{2}\cos \theta =2co{{s}^{2}}\theta$

$\Rightarrow 2{{\cos }^{2}}\theta -2\sqrt{2}\cos \theta -3=0$

$\cos \theta =\frac{2\sqrt{2}\pm \sqrt{{{\left( 2\sqrt{2} \right)}^{2}}-4\left( 3 \right)\left( 2 \right)}}{4}$

$\,\,\,\,\,\,\,\,\,=\frac{2\sqrt{2}\pm 4\sqrt{2}}{4}=\frac{\sqrt{2}\pm 2\sqrt{2}}{2}$

$\,\,\,\,\,\,\,=\frac{3\sqrt{2}}{2},\frac{-1}{\sqrt{2}}$

$\sin ce\,\,1\le \cos \theta \le -1$

$Hence\,\cos \theta =\frac{-1}{\sqrt{2}}$

$\theta =\frac{\pi }{4},\frac{7\pi }{4}\,\,\,\,\,\,\because \theta \in \left( 0,2\pi \right)$

$a)\,\cos 2\theta =1$

$\,\,\,\cos \left( 2\frac{\pi }{4} \right)$

$=\cos \left( \frac{\pi }{2} \right)$

$=0$

$b)\,sin2\theta =1$

$\,\,\,\sin \left( 2\frac{\pi }{4} \right)$

$=\sin \left( \frac{\pi }{2} \right)$

$=1$

$c)\,\tan \theta =1$

$\,\,\,\tan \left( \frac{\pi }{4} \right)$

$=1$

$d)\tan \theta =-1$

$\,\,\,\tan \left( \frac{7\pi }{4} \right)=\tan \left( 2\pi -\frac{\pi }{4} \right)$

$=-\tan \left( \frac{\pi }{4} \right)$

$=-1$

For

$x\ \in\ (0,\pi)$ , the equation

$\sin x+2\sin 2x-\sin 3x=3$ has

0%
infinitely solutions
0%
three solution
0%
one solution
0%
no solution

Explanation

$\sin x+2\sin 2x-\sin 3x=3$

$\sin x+4\sin x\cos x=3+\sin 3x$

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

$\sin x(1+4\cos x)=3+\sin x(3-4\sin^2 x)$

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

$4\cos^2x-4\cos x-2=9$

$(2\cos x-1)^2-3 \ge -3$

and on

$(0, \pi)$ we have

$0 < \sin x \le 1$ they

The

$LHS$ of your last equation. should be

$\ge -3$ on

$(0,\ \pi)$

No Solution in

$(0,\ \pi)$

The number of value of

$x\in (0,2\pi)$ satisfying

$\log_{\tan x}(2+4\cos^2x)=2$

0%
$\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{2}$
0%
$\dfrac{\pi}{3}$
0%
$\dfrac{\pi}{6}$

The sum

$s=sin\theta+sin2\theta+.......+sin n \theta$ , equals

0%
$sin\dfrac{1}{2}(n+1)\theta sin \dfrac{1}{2}n \theta/ sin \dfrac{\theta}{2}$
0%
$cos\dfrac{1}{2}(n+1)\theta sin \dfrac{1}{2}n \theta/ sin \dfrac{\theta}{2}$
0%
$sin\dfrac{1}{2}(n+1)\theta sin \dfrac{1}{2}n \theta/ cos \dfrac{\theta}{2}$
0%
$cos\dfrac{1}{2}(n+1)\theta sin \dfrac{1}{2}n \theta/ cos\dfrac{\theta}{2}$

$(1+\tan{A}).(1+\tan{B})=2$ , then

$A+B$ is

0%
$\dfrac{\pi}{2}$
0%
$\dfrac{\pi}{3}$
0%
$\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{6}$

Explanation

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

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

$(1)$

Now

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

$\therefore \tan A+\tan B=1-\tan A\tan B$

$\therefore\dfrac{\tan A+\tan B}{1-\tan A\tan B}=1$ ........

$(2)$

$\therefore$ From

$(1)$ &

$(2)$

$\tan (A+B)=1$

$\therefore A+B=\dfrac{\pi}{4}$

$3\sin { 2\theta } =2\sin { 3\theta }$ and

$0<\theta <\pi$ , then value of

$\sin { \theta }$ is

0%
$\dfrac{\sqrt{2}}{3}$
0%
` $\dfrac{\sqrt{3}}{\sqrt{5}}$
0%
$\dfrac{\sqrt{15}}{4}$
0%
$\dfrac{\sqrt{2}}{\sqrt{5}}$

Explanation

$3 \sin 2 \theta = 2 \sin 3 \theta$

$\rightarrow 3 \times 2 \sin \cos \theta = 2(3 \sin \theta - 4 \sin{3} \theta )$

$\rightarrow 6 \sin \theta \cos \theta = 6 \sin \theta - 8 \sin{3} \theta$

$\rightarrow \sin \theta (6 \cos \theta - 6+8 \sin^{2} \theta )=0$

$\therefore \sin \theta =0$ or

$3 \cos \theta+ 4 \sin{2} \theta =3$

$\therefore \sin \theta =0$ or

$3 \cos \theta + 4 (1- \cos{2} \theta)=3$

$\therefore \sin \theta =0$ or

$3 \cos \theta -4 \cos{2} \theta +1 =0$

$\therefore 4 \cos^{2} \theta- 3 \cos \theta -1 =0$

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

$\therefore 4 \cos \theta (\cos \theta -1) +1 (\cos \theta -1)=0$

$\therefore (4 \cos \theta +1) (\cos \theta -1)=0$

$\therefore \cos \theta = \dfrac{-1}{4}$ or

$\cos \theta =1$

$\therefore \sin \theta = \sqrt{1-\left( \dfrac{-1}{4} \right)^{2}}$

$\therefore \sin \theta =\sqrt{1- \dfrac{1}{16}}$

$\therefore \sin \theta = \dfrac{\sqrt{15}}{4}$

$\cos2x\ +\ 2\cos\ x\ =\ 1$ , then

$\sin^{2}x(2-\cos^{2}x)$ is

0%
$1$
0%
$2$
0%
$4$
0%
none of these

Explanation

Solution:- (A)

$1$

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

$2 \cos^{2}{x} - 1 + 2 \cos{x} = 1 \; \left( \because \cos{2x} = 2 \cos^{2}{x} - 1 \right)$

$\Rightarrow 2 \cos^{2}{x} + 2 \cos{x} - 2 = 0$

$\Rightarrow \cos^{2}{x} + \cos{x} = 1 ..... \left( 1 \right)$

$\Rightarrow 1 - \cos^{2}{x} = \cos{x}$

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

Now,

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

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

$= \cos{x} \left( 1 + \cos{x} \right) \; \left[ \text{From} \left( 2 \right) \right]$

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

$= 1 \; \left[ \text{From} \left( 1 \right) \right]$

Hence

$\sin^{2} \left( 2 - \cos^{2}{x} \right) = 1$ .

State whether the following statement is true or false:

$cosec{20^ \circ }\sec {20^ \circ } =$

0%
$2$
0%
$2 cosec {40^ \circ}$
0%
$4$
0%
$2\sin{45^ \circ} . cosec {40^ \circ}$

Explanation

$csc 20 \sec 20$

$\implies \dfrac 1{\sin 20\cos 20}$

$\implies \dfrac 2{2\sin 20\cos 20}$

$\implies \dfrac 2{\sin 2(20)}$

$\implies \dfrac 2{\sin 40}$

$\implies 2{\csc 40}$

${ cos }^{ 2 }2x+2{ cos }^{ 2 }x=1,x\epsilon \left( -\pi ,\pi \right)$ , then

$x$ can take the values

0%
$\pm \dfrac { \pi }{ 2 }$
0%
$\pm \dfrac { \pi }{ 4 }$
0%
$\pm \dfrac { 3\pi }{ 8 }$
0%
Non of these

Explanation

Given

$x\in (-\pi,\pi)\implies 2 x\in (-2\pi,2\pi)$

$\cos^2 2 x+2\cos^2 x=1$

$\implies \cos^2 2 x+2\cos^2 x-1=0$

$\implies \cos^2 2 x+(2\cos^2 x-1)=0$

$\implies \cos^2 2 x+\cos 2 x=0$

$\implies \cos 2 x(\cos 2 x+1)=0$

$\cos 2 x=0$ or

$-1$

$\implies 2 x=\pm \dfrac{\pi}{2},\pm \dfrac{3\pi}{2}$ or

$\pm \pi$

$\implies x=\pm \dfrac{\pi}{4},\pm \dfrac{3\pi}{4}$ or

$\pm\dfrac{\pi}{2}$

The number of solutions to the equation

$sin\ 5x+\ sin\ 3x\ +\ sin\ x=0$ for

$\ 0\leq x\leq \pi$ is

0%
1
0%
2
0%
3
0%
None of these

Explanation

$sin x+sin3x+sin5x = 0 x\in [0,\pi ]$

$(sinx+sin5x)+sin3x = 0$

$2sin 3x cos2x + sin3x = 0$

$\displaystyle [\because sinx+siny = 2\left | sin \dfrac{x+y}{2} \right | \left | cos\dfrac{x-y}{2} \right |]$

$sin 3x (2cos2x+1) = 0$

$\Rightarrow sin 3x = 0 , 2cos2x +1 = 0$

$3x = n\pi$

$\quad|$ ,

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

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

$\quad|$

$2x =2n\pi \pm \dfrac{2\pi }{3} [\because cos2\dfrac{2\pi }{3} = \dfrac{-1}{2}]$

$\quad \quad \quad \quad \, | x = n\pi \pm \dfrac{\pi }{3}$

For

$x\in [ 0,\pi ], x = 0, \dfrac{\pi }{3}, \dfrac{2\pi }{3}, \pi$ [ 4 solutions]

$\therefore$ None of these is correct option

State true or false.

${(\cos x + {\mathop{\rm cosy}\nolimits} )^2} + {(\sin x - \sin y)^2} = 4{\cos ^2}\left( {\frac{{x + y}}{2}} \right)$

0%
True
0%
False

Explanation

$(\cos x+\cos y)^2+(\sin x-\sin y)^2$

$=\left[\cos\left(\dfrac{x+y}{2}\right)\cos\left(\dfrac{x-y}{2}\right)\right]^2+\left[2\cos\left(\dfrac{x+y}{2}\right)\sin\left(\dfrac{x-y}{2}\right)\right]^2$

$=\left[(2)^2\cos^2\left(\dfrac{x+y}{2}\right)\cos^2\left(\dfrac{x-y}{2}\right)\right]+\left[(2)^2\cos^2\left(\dfrac{x+y}{2}\right)\sin^2\left(\dfrac{x-y}{2}\right)\right]$

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

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

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

$=4\cos^2\left(\dfrac{x+y}{2}\right)\times 1$

$=4\cos^2\left(\dfrac{x+y}{2}\right)$

$=R.H.S$

$\therefore$

$(\cos x+\cos y)^2+(\sin x-\sin y)^2$

$=4\cos^2\left(\dfrac{x+y}{2}\right)$

$\cos^{2}x=\sin x$ , then the value of

$\cos^{2}x(1+\cos^{2}x)$ is equal to

0%
$-1$
0%
$1$
0%
$0$
0%
$\dfrac{1}{2}$

Explanation

Given

$\cos^2 x=\sin x$

Squaring on both sides

$\implies (\cos^2 x)^2=\sin^2 x$

$\implies \cos^4 x=1-\cos^2 x$

$\implies \cos^4 x+\cos^2 x=1$

$\implies \cos^2 x(1+\cos^2 x)=1$

$\dfrac{{\sec 8{\text{A}} - 1}}{{\sec 4{\text{A}} - 1}} =$

0%
$\dfrac{{\tan 2{\text{A}}}}{{\tan 8{\text{A}}}}$
0%
$\dfrac{{\tan 8{\text{A}}}}{{\tan 2{\text{A}}}}$
0%
$\dfrac{{{\text{cot8A}}}}{{\cot 2{\text{A}}}}$
0%
$\dfrac{{{\text{tan6A}}}}{{\tan 2{\text{A}}}}$

Explanation

Solution :-

$\displaystyle \frac{sec8A-1}{sec4A-1} = \frac{1/cos8A-1}{1/cos4A-1}$

$\displaystyle \Rightarrow \frac{\frac{1-cos8A}{cos8A}}{\frac{1-cos4A}{cos4A}} = \frac{cos4A(1-cos8A)}{cos8A(1-cos4A)}$

$\displaystyle \Rightarrow \frac{cos4A(1-(1-2sin^{2}4A))}{cos8A(1-(1-2sin^{2}2A))} = \frac{cos4A.sin^{2}2A}{cos8A.sin^{2}2A}$

$\displaystyle \Rightarrow \frac{(2cos4Asin4A)sin4A}{(2cos8A\,sin^{2}2A)} = \frac{sin8Asin4A}{2cos8A.sin^{2}2A}$

$\displaystyle \Rightarrow tan8A(\frac{sin4A}{2sin^{2}A}) = tan8A.(\frac{2sin2A.cos2A}{2sin^{2}2A})$

$\displaystyle \Rightarrow \frac{tan8A}{tan2A}$

1106892_1182322_ans_31ca8445cb9442fb8309516b14a22992.jpg

$\tan \theta + \, \cot \theta = 4$ , then the value of

$\tan^3 \theta + \cot^3 \theta$ is

0%
$52$
0%
$16$
0%
$7 \dfrac{9}{8}$
0%
$27 \dfrac{1}{27}$

Explanation

$\tan\theta +\cot\theta =4$

$(\tan\theta +\cot\theta )^{3}=4^{3}$

$\tan^{3}\theta +\cot^{3}\theta +(\tan\theta +\cot\theta )3\tan\theta \cot\theta =64$

$\tan^{3}\theta +\cot^{3}\theta +(4)3=64$

$\tan^{3}\theta +\cot^{3}\theta =64-12$

$\tan^{3}\theta +\cot^{3}\theta =52$

The positive integer value of

$n>3$ satisfing the equation

$\frac{1}{{\sin \left( {\frac{\pi }{n}} \right)}} = \frac{1}{{\sin \left( {\frac{{2\pi }}{n}} \right)}} + \frac{1}{{\sin \left( {\frac{{3\pi }}{n}} \right)}}$
is

0%
$7$
0%
$8$
0%
$9$
0%
$10$

$cot\Theta =sin2\Theta (where\Theta \neq n\pi ,n\ is\ an\ integer)\Theta$ =

0%
45 $^{0}$ and 60 $^{0}$
0%
45 $^{0}$ and 90 $^{0}$
0%
only 45 $^{0}$
0%
only 90 $^{0}$

Explanation

$\cot\theta =\sin 2\theta$

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

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

$\Rightarrow \cos\theta -2\sin^2\theta \cos\theta =0$

$\Rightarrow \cos\theta(1-2\sin^2\theta)=0$

$\Rightarrow \cos\theta(\cos 2\theta)=0$

$\Rightarrow \cos\theta =0$ or

$\cos 2\theta =0$

$\Rightarrow \theta =\dfrac{\pi}{2}$ or

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

$\Rightarrow \theta =\dfrac{\pi}{4}$

$\therefore \theta$ can be

$45^o$ or

$90^o$ .

1188528_1195104_ans_d12945a11b54498ab408a5b5d55f707e.jpg

$(\sec\ A-\cos\ A)(\sec\ A+\cos\ A)=\sin^{2}\ A+\tan^{2}A$ .

0%
True
0%
False

Explanation

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

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

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

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

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

In a triangle ABC,

$b\,cos\,(C+\theta)+c\,cos(B-\theta)=$

0%
$a\,cos\,\theta$
0%
$a\,sin\,\theta$
0%
$a\,tan\,\theta$
0%
$a\,cot\,\theta$

Explanation

$\triangle ABC,A+B+C=\pi$ and

$a=2 R\sin A,b=2 R\sin B,c=2 R \sin C$

$b\cos (C+\theta)+c\cos (B-\theta)=R(2\sin B\cos (C+\theta)+2\sin C\cos (B-\theta))$

$=R(\sin (B+C+\theta)+\sin (B-C-\theta)+\sin (B+C-\theta)+\sin (C-B+\theta))$

$=R(\sin (\pi-(A-\theta))+\sin (\pi-(A+\theta))+\sin (B-C-\theta)-\sin (B-C-\theta))$

$=R(\sin (A-\theta)+\sin (A+\theta))$

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

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

$=a\cos \theta$

CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 10 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 10 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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