If $$\dfrac{\sin^4x}{2}+\dfrac{\cos^4x}{3}=\dfrac{1}{5}$$, then

$$\dfrac{\sin^8x}{8}+\dfrac{\cos^8x}{27}=\dfrac{1}{125}$$

$$\dfrac{\sin^8x}{8}+\dfrac{\cos^8x}{27}=\dfrac{2}{125}$$

MCQ Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 15

The solution set of the equation $4 \sin \theta-2 \cos \theta-2\sqrt {3} \sin \theta+\sqrt {3}=0$ in the interval $(0,2\pi)$ is-

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

Explanation

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

$\Rightarrow (2\cos\theta-\sqrt 3)(2\sin\theta-1)=0$

$\therefore \cos \theta=\cfrac{\sqrt 3}{2}$ or

$\sin\theta=\cfrac{1}{2}$

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

$\theta=\cfrac{\pi}{6}, \cfrac{11\pi}{6}, \cos \theta=\cfrac{\sqrt 3}{2}$

$\theta=\cfrac{\pi}{6}, \cfrac{5\pi}{6}, \sin \theta=\cfrac{1}{2}$

$\therefore \theta=\cfrac{\pi}{6}, \cfrac{5\pi}{6}, \cfrac{11\pi}{6}$

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

$\sin 2\theta$ is equal to:

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

Explanation

Given that

$\sin\theta+\cos\theta=1$

now squaring both side we have

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

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

$1+2\sin\theta \cos\theta=1$ (

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

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

$\sin2\theta=0$ (

$\sin2\theta=2\sin\theta \cos\theta$ )

hence option C is correct.

The least value of a for which the expression

$f(x)=\cfrac{4}{\sin x}+\cfrac{1}{1-\sin x}=a^{2}$ has at least one solution on the interval

$(0,\quad \pi /2)$ is

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

Explanation

Let

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

$f^{'}(x)=[\cfrac{-4}{\sin^{2}x}+\cfrac{1}{(1-\sin x)^{2}}]\cos x$

$f(x)$ is not monotonically increasing or decreasing function.

$\therefore a^{2}$ will be minimum , for

$x$ at which

$f^{'}(x)=0$

$\therefore \cfrac{-4}{\sin^{2}x}+\cfrac{1}{(1-\sin x)^{2}}=0$ (

$\cos x \ne 0)$ ,

$(x \in(0,\cfrac{\pi}{2}))$

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

$2-2\sin x=\sin x$

$\therefore \sin x=\cfrac{2}{3}$

$\therefore$ least value of

$a^{2}$ =4

$(\cfrac{3}{2})+\cfrac{1}{1-\cfrac{2}{3}}=6+3$

$\therefore a^{2}=9$

$\therefore$ least value of

$a=-3$

If x=

$\frac{{2\left( {\sin {1^o} + \sin {2^o} + \sin {3^o} + .... + \sin {{89}^o}} \right)}}{{2\left( {\cos {1^o} + \cos {2^o} + ..... + \cos {{44}^o}} \right) + 1}}$ , then the value of

${\log _x}2$ is equal

0%
1
0%
2
0%
1/2
0%
4

Total number of solutions of

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

$[-2\pi, 2\pi]$ is equal to:

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

The number of solution of

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

$\left(\dfrac{\pi}{2},\pi\right)$ is :

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

Simplifying $\sin 3x = \cos 2x$ .

$\sin 3x = \cos 2x$

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

$4{\sin ^3}x - 2{\sin ^2}x - 3\sin x + 1 = 0$

Put $\sin x = t$ .

$4{t^3} - 2{t^2} - 3t + 1 = 0$

$\left( {t - 1} \right)\left( {4{t^2} + 2t - 1} \right) = 0$

$t = 1$

Or,

$4{t^2} + 2t - 1 = 0$

$t = \frac{{ - 2 \pm \sqrt {{{\left( 2 \right)}^2} - 4\left( 4 \right)\left( { - 1} \right)} }}{{2 \times 4}}$

$= \frac{{ - 2 \pm \sqrt {20} }}{8}$

$= \frac{{ - 2 \pm 2\sqrt 5 }}{8}$

$= \frac{{ - 1 \pm \sqrt 5 }}{4}$

Then,

$\sin x = 1$

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

Or,

$\sin x = \frac{{ - 1 \pm \sqrt 5 }}{4}$

$\sin x = \frac{{ - 1 + \sqrt 5 }}{4}$

$x = \pi + \frac{\pi }{{10}}$

$x = \frac{{11\pi }}{{10}}$

Or,

$\sin x = \frac{{ - 1 - \sqrt 5 }}{4}$

$x = 2\pi - \frac{\pi }{{10}}$

$x = \frac{{19\pi }}{{10}}$

So, $x = \frac{\pi }{2}$ , $x = \frac{{11\pi }}{{10}}$ and $x = \frac{{19\pi }}{{10}}$ .

Since the given interval is $\left( {\frac{\pi }{2},\pi } \right)$ , then, $x = \frac{\pi }{2}$ is the only solution.

Therefore, the number of solutions of $\sin 3x = \cos 2x$ is 1.

$x\in \left[ 0,\dfrac { \pi }{ 2 } \right]$ , the number of solutions of the equation,

$\sin { 7x } +\sin { 4x } +\sin { x=0 }$ is

0%
$3$
0%
$5$
0%
$6$
0%
$None$

$\sin B=\dfrac {1}{5}\sin (2A+B)$ , then

$\dfrac {\tan (A+B)}{\tan A}$ is equal to

0%
$\dfrac53$
0%
$\dfrac23$
0%
$\dfrac32$
0%
$\dfrac35$

Explanation

$\sin B = \dfrac{1}{5} \sin (2A+B)$

$5 \sin B = \sin (2A+B)$

$5 \sin ((A+B)-A)= \sin ((A+B)+A)$

$5 [\sin (A+B) \cos A- \cos (A+B) \sin A]= \sin (A+B) \cos A + \cos (A ) \sin A$

$\Rightarrow 4 \sin (A+B) \cos A= 6 \cos (A+B) \sin A$

$\dfrac{\sin (A+B) . \cos A}{\cos (A+B). \sin A} = 6/ 4$

$\tan (A+ B) . \cot A = 3/2$

$\dfrac{\tan (A+B)}{\tan A}= 3/2$

Find the general solution of

$x$

$\cos^2 2x+\cos^2 3x=1$

0%
$(2k+1)\dfrac{\pi}{10},k \in I$
0%
$(\pi+1)\dfrac{\pi}{10};k \in I$
0%
$(2k-1)\dfrac{\pi}{10},k\in I$
0%
Both (A) and (C)

Explanation

general solution of

$\cos^22x+\cos^23x=1$

we know

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

$\boxed{\cos^2x=\frac{\cos 2x+1}{2}}$

So in the given question

$\left(\dfrac{\cos4x+1}{2}\right)+\left(\dfrac{\cos 6x+1}{2}\right)=1$

$\dfrac{\cos 4x}{2}+\dfrac{1}{2}+\dfrac{\cos 6x}{2}+\dfrac{1}{2}=1$

$\dfrac{\cos 4x}{2}=\dfrac{\cos 6x}{2}$

$\cos 4x=\cos (\pi+6x)$

we know

$\boxed{\cos x=cos y}$

then

$x=(2n\pi+y)$

$4x=(2k\pi\pm(\pi+6x))$

take

$+$

$-$

$4x=2k\pi+\pi+6x$

$-2x=2k\pi+\pi$

$x=-k\pi-\dfrac{\pi}{2}$

$4x=2k\pi-\pi-6x$

$10x=(2k-1)\pi$

$\boxed{x=(2k-1)\frac{\pi}{10}}$

The solution set of equation

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

$(0, 2\pi)$

0%
$\left\{ \dfrac { 3\pi }{ 4 } ,\dfrac { 7\pi }{ 4 } \right\}$
0%
$\left\{ \dfrac { \pi }{ 3 } ,\dfrac { 5\pi }{ 3 } \right\}$
0%
$\left\{ \dfrac { \pi }{ 6 } ,\dfrac { 5\pi }{ 6 } ,\dfrac{11\pi}{6}\right\}$
0%
None of these

Find the principal solution of the following equation:

$\tan { x=\sqrt { 3 } }$

0%
$x=\dfrac{\pi}{3}$ and $x=\dfrac{4}{3}\pi$
0%
$x=\dfrac{\pi}{2}$ and $x=\dfrac{1}{3}\pi$
0%
$x=\dfrac{\pi}{4}$ and $x=\dfrac{2}{3}\pi$
0%
$x={3}$ and $x=\dfrac{4}{3}\pi$

Explanation

We have,

$\tan {x}=\sqrt{3}$

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

$\boxed{x=\pi/3}------eq^n (1)$

Put again equation

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

$\boxed{x=\dfrac{4\pi}{3}}------eq^n (2)$

From equation

$(1)$ and

$(2)$

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

Hence this is the answer.

$\dfrac{\sin^4x}{2}+\dfrac{\cos^4x}{3}=\dfrac{1}{5}$ , then

0%
$\tan^2x=\dfrac{2}{3}$
0%
$\dfrac{\sin^8x}{8}+\dfrac{\cos^8x}{27}=\dfrac{1}{125}$
0%
$\tan^2x=\dfrac{1}{3}$
0%
$\dfrac{\sin^8x}{8}+\dfrac{\cos^8x}{27}=\dfrac{2}{125}$

The value of

$\quad ({ tan }^{ 2 }\cfrac { \pi }{ 7 } +{ tan }^{ 2 }\cfrac { 2\pi }{ 7 } +{ tan }^{ 2 }\cfrac { 3\pi }{ 7 } )$ x

$({ cot }^{ 2 }\cfrac { \pi }{ 7 } +cot^{ 2 }\cfrac { 2\pi }{ 7 } +{ cot }^{ 2 }\cfrac { 3\pi }{ 7 } )$ is

0%
$105$
0%
$35$
0%
$210$
0%
None of these

The number of real solutions

$x$ of the equation

$\cos ^ { 2 } ( x \sin ( 2 x ) ) + \frac { 1 } { 1 + x ^ { 2 } } - \cos ^ { 2 } x + \sec ^ { 2 } x$ is

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

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

0%
$\tan\left(\frac{\pi}{4}+\frac{A}{2}\right)$
0%
$\tan\left(\frac{\pi}{2}+\frac{A}{2}\right)$
0%
$\cot\left(\frac{\pi}{4}+\frac{A}{2}\right)$
0%
None of these

Explanation

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

Dividing numerator and denoinator by

$\cos\cfrac{A}{2}$

$\cfrac { 1+\tan\cfrac { A }{ 2 } }{ 1-\tan\cfrac { A }{ 2 } } =\cfrac { \tan\cfrac { \pi }{ 4 } +\tan\cfrac { A }{ 2 } }{ 1-\tan\cfrac { \pi }{ 2 } .\tan\cfrac { A }{ 2 } } =\tan\left( \cfrac { \pi }{ 4 } +\cfrac { A }{ 2 } \right)$

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

0%
True
0%
False

General solution of

${\sin ^3}x + {\cos ^3}x + \frac{3}{2}\sin 2x = 1$

0%
$x = n\pi$ when n is even integer
0%
$x = 2n\pi$ when n is odd integer
0%
$x = n\pi + \frac{\pi }{2}$ when n is odd integer
0%
$x = n\pi - \frac{\pi }{2}$

Explanation

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

$\Rightarrow\,{\sin}^{3}{x}+{\cos}^{3}{x}+\dfrac{3}{2}\times 2\sin{x}\cos{x}=1$

$\Rightarrow\,{\sin}^{3}{x}+{\cos}^{3}{x}+3\sin{x}\cos{x}\left(1\right)$ since

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

$\Rightarrow\,{\sin}^{3}{x}+{\cos}^{3}{x}+3\sin{x}\cos{x}\left({\sin}^{2}{x}+{\cos}^{2}{x}\right)=1$

$\Rightarrow\,{\left(\sin{x}+\cos{x}\right)}^{3}=1$

$\Rightarrow\,\sin{x}+\cos{x}=1$

$\Rightarrow\,\dfrac{1}{\sqrt{2}}\sin{x}+\dfrac{1}{\sqrt{2}}\cos{x}=\dfrac{1}{\sqrt{2}}$

$\Rightarrow\,\cos{\dfrac{\pi}{4}}\sin{x}+\sin{\dfrac{\pi}{4}}\cos{x}=\dfrac{1}{\sqrt{2}}$

$\Rightarrow\,\sin{\left(x+\dfrac{\pi}{4}\right)}=\sin{\dfrac{\pi}{4}}$

$\Rightarrow\,x+\dfrac{\pi}{4}=n\pi+\dfrac{\pi}{4}$

$\Rightarrow\,x=n\pi$ where

$n$ is an even integer.

The number of solution of

$\sec x \cos 5x+1=0$ in the interval

$[0,2\pi]$ is

0%
$5$
0%
$8$
0%
$10$
0%
$12$

$2 \tan ^ { 2 } \theta - 5 \sec \theta = 1$ has exactly

$7$ solutions in the interval

$\left[ 0 , \dfrac { n \pi } { 2 } \right] , n \in N$ then the least and greatest values of

$n$ are

0%
$6 , 8$
0%
$12 , 14$
0%
$13 , 15$
0%
$15 , 17$

The value of

$\sqrt { 3 } \cot 20 ^ { \circ } - 4 \cos 20 ^ { \circ }$ = __________________________.

0%
$1$
0%
$-1$
0%
$0$
0%
none of these

If in a triangle

$ABC, \dfrac{b+c}{11}=\dfrac{c+a}{12}=\dfrac{a+b}{13}$ , then

$\cos A$ is equal to:

0%
$\dfrac{19}{35}$
0%
$\dfrac{5}{7}$
0%
$\dfrac{1}{5}$
0%
$\dfrac{35}{19}$

Explanation

$\dfrac{b+c}{11}=\dfrac{c+a}{12}=\dfrac{a+b}{13}=k$ .

$b+c=11\ k,\, c+a=12\ k, \,a+b=13\ k$ .

$2(a+b+c)=36\ k$

$a+b+c=18\ k$

$a=7\ k, b=6\ k, c=5\ k$

$\cos A=\dfrac{b^2+c^2-a^2}{2bc}$

$=\dfrac{(6k)^2+(5k)^2-(7k)^2}{2\times 6k\times 5k}$

$=\dfrac{61\ k^2-49\ k^2}{60\ k^2}=\dfrac{12\ k^2}{60\ k^2}$

$\cos A=\dfrac{1}{5}$

The number of real solutions of the equation

$\sin \left( e ^ { x } \right) = 5 ^ { x } + 5 ^ { - x }$ is ____________________.

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

The value of

$\sin \dfrac {2\pi}{7}+\sin \dfrac {4\pi}{7}+\sin \dfrac {8\pi}{7}$ is

0%
$1$
0%
$\dfrac {\sqrt {7}}{2}$
0%
$\dfrac {3\sqrt {3}}{4}$
0%
$\dfrac {\sqrt {15}}{4}$

The value of

$\tan\theta+\sec\theta$ is equal to

0%
$3t$
0%
$t$
0%
$t-t^{2}$
0%
$t^{2}-2t$

$\tan^{2}\alpha-\tan^{2}\beta-\dfrac{1}{2}\sin(\alpha-\beta)\sec^{2}\alpha\sec^{2}\beta$ is zero then value of

$\sin(\alpha+\beta)$ is

$(\alpha\neq\beta)$

0%
$1$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{2}$

The value of

$\sin{\left(\alpha+\beta\right)}$ is

0%
$\dfrac{24}{25}$
0%
$\dfrac{12}{13}$
0%
$\dfrac{1}{25}$
0%
$None\ of\ these$

In a

$\triangle ABC, \sum a \cos A=4 \sin A \sin B \sin C$ , then value of

$\left(\dfrac{\sum \sin A}{\sum a}\right)^{2}$ is-

0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{2}$
0%
$1$
0%
$\dfrac{1}{3}$

$sin\left( { 4tan }^{ -1 }{ \cfrac { 1 }{ 3 } } \right) =$

0%
$\frac { 12 }{ 25 }$
0%
$\frac { 24 }{ 25 }$
0%
$\frac { 1 }{ 5 }$
0%
none of these

$0 < \phi < \dfrac {\pi}{2}$ and if

$\displaystyle x=\sum^{\alpha}_{n=0}\cos^{2}\phi:y=\sum^{\alpha}_{n=0}\sin^{2n}\phi$ then

0%
$x^{2}+y^{2}=1$
0%
$x^{2}-y^{2}=1$
0%
$x+y=xy$
0%
$\dfrac {1}{x}+\dfrac {1}{y}=1$

${ tan }^{ 2 }\theta +{ cot }^{ 2 }\theta =\frac { a }{ b-cos4\theta } +c$ (where defined) is satisfied for all

$\theta$ , then the value of (a + b + c) is equal to

$(a,b,c\in I)$

0%
6
0%
7
0%
8
0%
9

CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 15 - 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 15 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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