MCQ Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 1

The number of

$x\epsilon [0, 2\pi]$ for which

$|\sqrt {2\sin^{4} x + 18\cos^{2}x} - \sqrt {2\cos^{4} x + 18\sin^{2} x}| = 1$ is:

0%
$4$
0%
$2$
0%
$6$
0%
$8$

Explanation

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

$sin^4x+cos^4x+2sin^2xcos^2x=1$

$sin^4x+cos^4x=1-2sin^2xcos^2x$ ...[1]

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

$sin^6x+cos^6x+3sin^2xcos^2x(sin^2x+cos^2x)=1$

$sin^6x+cos^6x=1-3sin^2xcos^2x$ ...[2]

$|\sqrt{2sin^4x+18cos^2x}-\sqrt{2cos^4x+18sin^2x}|=1$

Squaring above equation

$\therefore 2sin^4x+18cos^2x+2cos^4x+18sin^2x-2(\sqrt{2sin^4x+18cos^2x})(\sqrt{2cos^4x+18sin^2x})=1$

$\therefore 2(sin^4x+cos^4x)+18(cos^2x+sin^2x)-1=2(\sqrt{2sin^4x+18cos^2x})(\sqrt{2cos^4x+18sin^2x})$

$\therefore 2(1-2sin^2xcos^2x)+18-1=2(\sqrt{2sin^4x+18cos^2x})(\sqrt{2cos^4x+18sin^2x})$ ..( [Using[1] )

$\therefore 19-4sin^2xcos^2x=2(\sqrt{2sin^4x+18cos^2x})(\sqrt{2cos^4x+18sin^2x})$

Squaring above equation

$\therefore 361+16sin^4xcos^4x-152sin^2xcos^2x=4(4sin^4xcos^4x+36cos^6x+36sin^6x+324sin^2xcos^2x)$

$\therefore 361+16sin^4xcos^4x-152sin^2xcos^2x=16sin^4xcos^4x+144cos^6x+144sin^6x+1296sin^2xcos^2x$

$\therefore 361=144(cos^6x+sin^6x)+1448sin^2xcos^2x$

$\therefore 361=144(1-3sin^2xcos^2x)+1448sin^2xcos^2x$ ....( Using [2] )

$\therefore 217=1016sin^2xcos^2x$

$\therefore 217=254sin^22x$

$\therefore sin2x=\pm \sqrt{\dfrac{217}{254}}$

Let

$sin^{-1} \sqrt{\dfrac{217}{254}}=\theta$

$sin2x=\pm \sqrt{\dfrac{217}{254}}$

$\therefore 2x=sin^{-1} \sqrt{\dfrac{217}{254}}$

$\therefore 2x=\theta,\pi-\theta,2\pi+\theta,3\pi-theta$

$\therefore x=\dfrac{\theta}{2},\dfrac{\pi-\theta}{2},\dfrac{2\pi+\theta}{2},\dfrac{3\pi-\theta}{2}$

$2x=sin^{-1} (-\sqrt{\dfrac{217}{254}})$

$\therefore 2x=\pi+\theta,2\pi-\theta,3\pi+\theta,4\pi-\theta$

$\therefore x=\dfrac{\pi+\theta}{2},\dfrac{2\pi-\theta}{2},\dfrac{3\pi+\theta}{2},\dfrac{4\pi-\theta}{2}$

So, total number of solutions =8

So, the answer is option (D)

For

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

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

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

Explanation

$\sin x + 2 \sin2x - \sin 3 x = 3$

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

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

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

$\sin x [3 - (2 \cos x - 1)^2] =3$

$\Rightarrow \sin x = 1$ and $2 \cos x - 1 =0$

$\Rightarrow x = \dfrac{\pi}{2}$ and $x = \dfrac{\pi}{3}$

Which is not possible at same time.

Hence, no solution.

$0^{\circ}\leq \theta\leq 90^{\circ}$ and

$\sqrt{3} tan\theta - sec\theta=1$ , then

$\theta$ has the value

0%
$30^{\circ}$
0%
$45^{\circ}$
0%
$60^{\circ}$
0%
$90^{\circ}$

Explanation

$\sqrt{3} tan \theta -1 =sec \theta$
Squaring and substituting

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

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

$\therefore 2 tan \theta (tan \theta - \sqrt{3}) =0$

$\theta = 60^{\circ}$ is one of the values.

The of value of

$\quad \sin60^{\small\circ}\cos30^{\small\circ} + \cos60^{\small\circ}\sin30^{\small\circ}$ is equal to

0%
$\sin 90^0$
0%
$\sin 45^0$
0%
$\sin 180^0$
0%
$\sin 30^0$

Explanation

$sin60^{0}cos30^{0}+cos60^{0}sin30^{0}$

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

$=1$

$=sin90^{0}$

$4cos^{3}30^{0}-3cos30^{0}$

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

$=0$

$=cos90^{0}$

Which one of the following relations is true ?

0%
$\cos^2 \theta - \sin^2 \theta=1$
0%
$\text{cosec}^2 \theta - \sec^2 \theta=1$
0%
$\cot^2 \theta - \tan^2 \theta=1$
0%
$\sec^2 \theta - \tan^2 \theta=1$

Explanation

(A) L.H.S

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

$=1-2 { \sin }^{ 2 }\theta$ is never

$=1$ .

So this statement is not correct.

(B) L.H.S

$={ \text{cosec} }^{ 2 }\theta -{ \sec }^{ 2 }\theta$

$=\dfrac { 1 }{ { \cos }^{ 2 }\theta } -\dfrac { 1 }{ { \sin }^{ 2 }\theta } =\dfrac { { \sin }^{ 2 }\theta -{ \cos }^{ 2 }\theta }{ { \sin }^{ 2 }\theta } \neq 1\neq R.H.S$

$\therefore$ This statement is not correct.

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

$=\dfrac { { \cos }^{ 2 }\theta }{ { \sin }^{ 2 }\theta } -\dfrac { { \sin }^{ 2 }\theta }{ { \cot }^{ 2 }\theta } =\dfrac { { \cos }^{ 4 }\theta -{ { \sin }^{ 4 } }\theta }{ { \sin }^{ 2 }\theta { \cos }^{ 2 }\theta }$

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

$=\dfrac { { \cos }^{ 2 }\theta -{ \sin }^{ 2 }\theta }{ { \sin }^{ 2 }\theta { \cos }^{ 2 }\theta } \neq 1\neq R.H.S$

So this statement is not correct.

(D) R.H.S

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

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

This statement is correct.

Which is true for all values of

$\theta$ ?

0%
$\sin^2 \theta -\cos^2\theta=1$
0%
$\text{cosec}^2\theta- \cot^2 \theta=1$
0%
$\text{cosec}^2 \theta - \cos^2 \theta=1$
0%
$\sec^2 \theta - \sin^2 \theta=1$

Explanation

We need to verify every option

$(A)$

Consider,

$\sin^2 \theta - \cos^2 \theta$

$=1-\cos^2 \theta-\cos^2 \theta$

$=1-2\cos^2 \theta$

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

Now,

$-\cos 2\theta=1$ if

$\cos 2\theta=-1\implies 2\theta=\cos^{-1}(-1)=(2n+1)\pi$

$\implies \theta=\dfrac{(2n+1)\pi}{2}, n\in \mathbb{Z}$

Thus, it is not true for every

$\theta$

$(B)$

Consider,

$\text{cosec}^2\theta-\cot^2\theta$

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

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

$\therefore \text{cosec}^2\theta-\cot^2\theta=1,$ For every value of

$\theta$

$(C)$

$\text{cosec}^2\theta-\cos^2 \theta=1\implies 1-\sin^2\theta \cos^2\theta=\sin^2\theta$

$\implies 1-\sin^2\theta=\sin^2\theta \cos^2\theta$

$\implies \sin^2\theta=1 \implies \theta=\dfrac{(2n+1)\pi}{2}$

Thus,

$\text{cosec}^2\theta-\cos^2 \theta=1$ is true for

$\theta=\dfrac{(2n+1)\pi}{2}$ but not for all values of

$\theta$ .

$(D)$

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

$\implies1-\sin^2\theta \cos^2\theta=\cos^2\theta$

$\implies \sin^2\theta=\sin^2\theta \cos^2\theta \implies \cos^2\theta=1$

$\implies \cos \theta=\pm 1 \implies \theta=n\pi, n\in \mathbb{Z}$

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

$\theta\in \mathbb{Z}$ but not for all values of

$\theta$ .

Hence, option B is correct.

$0^{\circ} \leq \theta \leq 90^{\circ}$ and

$\sqrt{2} tan \theta - sec \theta =0$ , then the value of

$(\sqrt{2} sin \theta + 2 tan \theta)$ is -

0%
$\sqrt{2}+2$
0%
$\frac{\sqrt{2}}{2} + \frac{2}{\sqrt{3}}$
0%
3
0%
2

Explanation

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

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

$\sqrt{2} \sin \theta + 2 \tan \theta = \sqrt{2} \times \frac{1}{\sqrt{2}} + 2 \times 1 =3$

The value of

$1+ \cot^2 A$ is

0%
$\cos^2 A$
0%
$\sec^2 A$
0%
$\tan^2 A$
0%
$\text{cosec}^2 A$

Explanation

We know that,

$\cot A= \dfrac{\text{base}}{\text{perpendicular}} = \dfrac bp$

$1+\cot^2A=1+\dfrac{b^2}{p^2}=\dfrac{b^2+p^2}{p^2}=\dfrac{h^2}{p^2}=\text{cosec}^2A$

$sin^2x+cos^2x=$

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

Explanation

Identity used:

$\sin^2\theta+\cos^2\theta=1$
Option C is correct.

The solution of equation

$\cos^2 \theta +\sin \theta+1=0$ lies in the interval

0%
$\displaystyle \left ( - \frac{\pi}{4}, \frac{\pi}{4} \right )$
0%
$\displaystyle \left ( \frac{\pi}{4}, \frac{3\pi}{4} \right )$
0%
$\displaystyle \left ( \frac{3\pi}{4}, \frac{5\pi}{4} \right )$
0%
$\displaystyle \left ( \frac{5\pi}{4}, \frac{7\pi}{4} \right )$

Explanation

$cos^2 \theta + \sin \theta + 1= 0$

$\Rightarrow 1 - \sin ^2 \theta + \sin \theta + 1 = 0$

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

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

$-1 < \sin \theta < 1$ so

$\sin \theta + 1 = 0$

$\Rightarrow \sin \theta = \sin \displaystyle \frac{3 \pi}{2}$

$\Rightarrow \theta = \displaystyle \frac{3\pi}{2} \varepsilon \left( \frac{5 \pi}{4}, \frac{7 \pi}{4} \right )$

The value of

$5\tan ^{ 2 }{ \theta } -5\sec ^{ 2 }{ \theta }$ is :

0%
$1$
0%
$-5$
0%
$0$
0%
$5$

Explanation

$5{ \tan }^{ 2 }\theta -5{ \sec }^{ 2 }\theta \\ =5({ \tan }^{ 2 }\theta- { \sec }^{ 2 }\theta )$

We know that

${ \tan }^{ 2 }\theta +1={ \sec }^{ 2 }\theta \\ \Rightarrow { \tan }^{ 2 }\theta -{ \sec }^{ 2 }\theta =-1\\ \therefore 5(-1)=-5$

$\displaystyle \frac { \sin { A } }{ 1+\cos { A } } +\frac { \sin { A } }{ 1-\cos { A } }$ is equal to :

0%
$\displaystyle \sin { A }$
0%
$\displaystyle 2cosecA$
0%
$\displaystyle \cos { A }$
0%
None of these

Explanation

$\displaystyle \frac { \sin { A } }{ 1+\cos { A } } +\frac { \sin { A } }{ 1-\cos { A } }$

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

$\displaystyle =\sin { A } \left( \frac { 2 }{ { sin }^{ 2 }A } \right) =\frac { 2 }{ \sin { A } } =2cosecA$

The minimum value of

$\displaystyle \sec ^{2}\alpha + \cos ^{2}\alpha$ is

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

Explanation

Consider the given expression,

${{\sec }^{2}}\alpha +{{\cos }^{2}}\alpha$

AS it is clear that ${{\cos }^{2}}\alpha$ and ${{\sec }^{2}}\alpha$ are positive numbers

So, we can apply the theorem of A.M and G.M

A.M>=G.M

so we have

$\left( \dfrac{{{\sec }^{2}}\alpha +{{\cos }^{2}}\alpha }{2} \right)=\sqrt{{{\sec }^{2}}\alpha +{{\cos }^{2}}\alpha }$

Or ${{\sec }^{2}}\alpha +{{\cos }^{2}}\alpha =2\sqrt{{{\sec }^{2}}\alpha +{{\cos }^{2}}\alpha }=2.1$

${{\sec }^{2}}\alpha +{{\cos }^{2}}\alpha =2$

So, minimum value of ${{\sec }^{2}}\alpha +{{\cos }^{2}}\alpha$ is 2.

$\cfrac { \tan { \theta } }{ \sec { \theta } -1 } +\cfrac { \tan { \theta } }{ \sec { \theta } +1 }$ is equal to

0%
$\text{cosec} { \theta }$
0%
$2\text{cosec} { \theta }$
0%
$\sec { \theta }$
0%
$2\sec { \theta }$

Explanation

$\cfrac { \tan { \theta } \left( \sec { \theta } +1+\sec { \theta } -1 \right) }{ \sec ^{ 2 }{ \theta } -1 } =\cfrac { 2\tan { \theta } \sec { \theta } }{ \sec ^{ 2 }{ \theta }-1 } =\cfrac { 2\tan { \theta } \sec { \theta } }{ \tan ^{ 2 }{ \theta } }$

$\Rightarrow \cfrac { 2\sec { \theta } }{ \tan { \theta } } =2\cfrac { 1 }{ \cos { \theta } .\cfrac { \sin { \theta } }{ \cos { \theta } } } =2\text{cosec} { \theta }$

If $\sin { \theta } +\cos { \theta } =p$ and $\tan { \theta } +\cot { \theta } =q$ , then $q\left( { p }^{ 2 }-1 \right) =$

0%
$\dfrac { 1 }{ 2 }$
0%
$2$
0%
$1$
0%
$3$

Explanation

Let

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

$\Rightarrow 1+\sin { 2\theta } ={ p }^{ 2 }$

Also

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

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

$\csc { 2\theta } =\dfrac { q }{ 2 }$

$\Rightarrow \dfrac { 1 }{ { p }^{ 2 }-1 } =\dfrac { q }{ 2 }$

$\Rightarrow \left( { p }^{ 2 }-1 \right) q\ =\ 2$

$\cfrac { 1 }{ \sec { \theta } -\tan { \theta } }$ is equal to

0%
$\sec { \theta } -\tan { \theta }$
0%
$\sec { \theta } \tan { \theta }$
0%
$\sec { \theta } +\tan { \theta }$
0%
$\sec { \theta }$

Explanation

$\cfrac { 1 }{ \sec { \theta } -\tan { \theta } } \times \cfrac { \sec { \theta } +\tan { \theta } }{ \sec { \theta } +\tan { \theta } } =\cfrac { \sec { \theta } -\tan { \theta } }{ \sin ^{ 2 }{ \theta } -\tan ^{ 2 }{ \theta } } =\sec { \theta } +\tan { \theta } \quad \quad$

$\tan\theta +\cot\theta =2$ , then the value of

$\tan^2\theta +\cot^2\theta$ is __________?

0%
$4$
0%
$2$
0%
$\displaystyle\frac{3}{2}$
0%
$5$

Explanation

Given,

$\tan \theta+\cot \theta=2$

Squaring both sides, we get

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

$\Rightarrow \tan^2\theta+\cot^2\theta+2=4$ .....Since

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

$\Rightarrow \tan^2\theta+\cot^2\theta=2$

$\cos ^{ 4 }{ A } -\sin ^{ 4 }{ A }$ is equal to

0%
$\sin { 2A }$
0%
$-\sin { 2A }$
0%
$\cos { 2A }$
0%
$-\cos { 2A }$

Explanation

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

What is

$\tan ( 360^o - A)$ ?

0%
$\tan A$
0%
$-\tan A$
0%
$\cot A$
0%
$-\cot A$

Explanation

$360 ^{°} - A$ lies in fourth quadrant and

$\tan$ is negative in fourth quadrant

$\Rightarrow \tan (360^{°} - A) =-\tan A$

Value of

$\theta (0 < \theta < 360^o )$ which satisfy the equation

$\text{cosec }\theta +2 =0$ is:

0%
$210^o, 100^o$
0%
$240^o, 300^o$
0%
$210^o, 240^o$
0%
$210^o, 330^o$

Explanation

Given:

$\text{cosec }\theta +2 =0$

$\Rightarrow \sin \theta =-\dfrac {1}{2}$

$\Rightarrow \theta =\left (\pi +\dfrac {\pi} {6}\right),$

$\left(2\pi - \dfrac {\pi} {6}\right)$

$\Rightarrow \theta=\dfrac{7\pi}{6},\dfrac{11\pi}{6}$

$\Rightarrow \theta =210^{°},\,330^{°}$

${\tan ^2}\theta = 2\,{\tan ^2}\phi + 1$ , then the value of

$\cos \,2\theta + {\sin ^2}\phi \,is$ is

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

Explanation

${\tan ^2}\theta = 2{\tan ^2}\phi + 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \left( 1 \right)$

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

$\dfrac{{1 - \left( {2{{\tan }^2}\phi + 1} \right)}}{{1 + {{\tan }^2}\theta }}$

$\dfrac{{ - 2{{\tan }^2}\phi }}{{1 + \left( {2{{\tan }^2}\phi + 1} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,by\,\left( 1 \right)$

$\dfrac{{ - 2{{\tan }^2}\phi }}{{2{{\sec }^2}\phi }}$

$- {\sin ^2}\phi$

$\cos 2\theta + {\sin ^2}\phi = 0$

$(1 - \cos A)/2 = x$ , then the value of

$x$ is

0%
$co{s^2}(A/2)$
0%
$\sqrt {\sin (A/2)}$
0%
$\sqrt {\cos (A/2)}$
0%
${\sin ^2}(A/2)$

Explanation

${\textbf{Step 1: Solve the given problem using the rules}}{\textbf{.}}$

$x = \dfrac{{1 - \cos A}}{2}$

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

$\boldsymbol{\mathbf{\left[ {\because \cos A = 1 - 2{{\sin }^2}\dfrac{A}{2}} \right]}}$

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

$\Rightarrow x = {\sin ^2}\dfrac{A}{2}$

${\textbf{Hence, the value of }}\mathbf{x}{\textbf{ is si}}{{\textbf{n}}^2}\mathbf{\dfrac{A}{2}. }$

The solution set of the equation

$4\sin \theta \cos \theta - 2\cos \theta - 2\sqrt 3 \sin \theta + \sqrt 3 = 0$ in the interval

$(0,2\pi )$ is-

0%
$\left\{ {\frac{{3\pi }}{4},\frac{{7\pi }}{4}} \right\}$
0%
$\left\{ {\frac{\pi }{3},\frac{{5\pi }}{3}} \right\}$
0%
$\left\{ {\frac{\pi }{3},\frac{{5\pi }}{3},\frac{{3\pi }}{4},\frac{{7\pi }}{4}} \right\}$
0%
$\left\{ {\frac{\pi }{6},\frac{{5\pi }}{6},\frac{{11\pi }}{6}} \right\}$

Explanation

$\begin{array}{l}4\sin \theta cos\theta - 2cos\theta - 2\sqrt {3\sin \theta } + \sqrt 3 = 0\\ = 2\cos \theta (2\sin \theta - 1) - \sqrt 3 \left( {2\sin \theta - 1} \right) = 0\\ = (2\cos \theta - \sqrt 3 )(2\sin \theta - 1) = 0\\ = 2\cos \theta - \sqrt 3 = 0\\ = \cos \theta = \frac{{\sqrt 3 }}{2}\end{array}$

$= \theta = \frac{\pi }{6},\frac{{11\pi }}{6}$
Same as,

$\begin{array}{l}2\sin \theta = 1\\ = \sin \theta = \frac{1}{2}\\ = \theta = \frac{\pi }{6},\frac{{5\pi }}{6}\end{array}$
Therefore:-

$\theta = \frac{\pi }{6},\frac{{5\pi }}{6},\frac{{11\pi }}{6}$

The solution of

$\frac{{3\sin \theta - \sin 3\theta }}{{1 + \cos \theta }} + \frac{{3\cos \theta + \cos 3\theta }}{{1 - \sin \theta }} = 4\sqrt 2 \cos \left( {\theta + \frac{\pi }{4}} \right)$

0%
$n\pi$
0%
$n\pi + \frac{\pi }{{12}}$
0%
$n\pi \pm \frac{\pi }{2}$
0%
$2n\pi$

Explanation

$\sin3\theta = 3\sin\theta - 4\sin^3\theta$

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

$\implies \cfrac{4\sin^3\theta}{1+\cos\theta} + \cfrac{4\cos^3\theta}{1-\sin\theta}$

$\implies \cfrac{4\sin\theta ( 1-\cos^2\theta)}{1+\cos\theta}+ \cfrac{4\cos\theta (1-\sin^2\theta)}{1-\sin\theta}$

$\implies 4\sin\theta(1-\cos\theta) + 4\cos\theta(1+\sin\theta)$

$\implies 4(\sin\theta + \cos\theta)$

$\implies 4\sqrt2 (\cos(\theta-\cfrac{\Pi}{4}))$

Equating with RHS, we get

$\theta = n\pi$

A minimum value of

$\sin{x}\cos{2x}$ is-

0%
$1$
0%
$-1$
0%
$-2/3\sqrt{6}$
0%
None of these

Explanation

$\cos x\cos 2x=2\sin x\sin 2x\implies \cos x=3\cos 3x\implies \sin x=\pm\displaystyle \frac{1}{\sqrt{6}}\implies$ the minimum value of

$\sin x\cos 2x$ is

$\displaystyle \frac{-2}{3\sqrt{6}}$

$\tan\theta +\tan 4\theta +\tan 7\theta =\tan \theta \tan 4\theta \tan 7\theta$ , then the general solution is?

0%
$\theta =\dfrac{n\pi}{4}$
0%
$\theta =\dfrac{n\pi}{12}$
0%
$\theta =\dfrac{n\pi}{6}$
0%
None of these

Explanation

we know, that

$tanA+tanB+tanC = tanA.tanB.tanC$ if

$A+B+C = n\pi$

Hence,

$12\theta = n\pi$

$\theta = \cfrac{n\pi}{12}$

$\cot\left( {a + \beta } \right) = 0,$ then

$\sin\left( {a + 2\beta } \right)$ can be

0%
$-\sin a$
0%
$\sin \beta$
0%
$\cos a$
0%
$\cos \beta$

Explanation

Since

$\cot(α+β) = 0 ⇒ \cot(α+β) = \cot\ 90^\circ$
Therefore,

$(α + β) = 90^\circ$ ......(1)

Since,

$\sin(α+2β)$

$= \sin[(α + β) + β ]$

$= \sin[90°+ β ]$

$= \cos β$

The solutions of the equation

$\sin x+3\sin 2x+\sin 3x=\cos x+3\cos 2x+\cos 3x$ in the interval

$0\le x\le 2\pi$ , are

0%
$\dfrac{\pi}{8}, \dfrac{5\pi}{8}, \dfrac{2\pi}{3}$
0%
$\dfrac{\pi}{8}, \dfrac{5\pi}{8}, \dfrac{9\pi}{8}, \dfrac{13\pi}{8}$
0%
$\dfrac{4\pi}{3}, \dfrac{9\pi}{3}, \dfrac{2\pi}{3}, \dfrac{13\pi}{8}$
0%
$\dfrac{\pi}{8}, \dfrac{5\pi}{8}, \dfrac{9\pi}{3}, \dfrac{4\pi}{3}$

Explanation

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

$formula\,used$

$\sin C+\sin D=2\sin \frac{C+D}{2}\cos \frac{C-D}{2}$

$\cos C+\cos D=2\cos \frac{C+D}{2}\cos \frac{C-D}{2}$

$\cos \left( -\theta \right)=\cos \theta$

$\Rightarrow 2\sin 2x\cos x+3\sin 2x=2\cos 2x\cos x+3\cos 2x$

$\Rightarrow \sin 2x\left[ 2\cos x+3 \right]=\cos 2x\left[ 2\cos x+3 \right]$

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

$\Rightarrow \tan 2x=1=\tan \frac{\pi }{4}$

$\therefore 2x=n\pi +\frac{\pi }{4}$

$x=\frac{n\pi }{2}+\frac{\pi }{8}$

$0\le x\le 2\pi$

$\therefore \,x=\frac{\pi }{8},\frac{5\pi }{8},\frac{9\pi }{8},\frac{13\pi}8$

$X\sin { \left( { 90 }^{ \circ }-\theta \right) \cot { \left( { 90 }^{ \circ }-\theta \right) } } =\cos { \left( { 9 }0^{ \circ }-\theta \right) }$ , then x =

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0
0%
1
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-1
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2

Explanation

$x\sin(90-\theta)\cot(90^o-\theta)=\cos(90-\theta)$

$\Rightarrow x\cos \theta\tan\theta =\sin\theta$

$\Rightarrow x=1$ .

1194718_1158077_ans_18a5a446a85a41188ce7ae5888af64e6.jpg

Select write the most appropriate answer from the given alternative in each following questions:-
The principal solution of tan

$x=-\sqrt{3}$ are:

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$\frac{-\pi }{3}and \frac{2\pi }{3}$
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$\frac{\pi }{3}and \frac{5\pi }{3}$
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$\frac{2\pi }{3}and \frac{5\pi }{3}$
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$\frac{2\pi }{3}and \frac{4\pi }{3}$

Explanation

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

$general\,solution$

$x=n\pi -\frac{\pi }{3}$

$principal\,solution$

$for\,n=0$

$x=-\frac{\pi }{3}$

$for\,n=1$

$x=\frac{2\pi }{3}$

$Hence\,principal\,solution\,is:-\frac{\pi }{3}and\frac{2\pi }{3}$

CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 1 - 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 1 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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