MCQ Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 7

$\cos ^{ -1 }{ \left( \cfrac { p }{ a } \right) } +\cos ^{ -1 }{ \left( \cfrac { p }{ b } \right) } =\alpha$ , then

$\cfrac { { p }^{ 2 } }{ { a }^{ 2 } } +k\cos { \alpha } +\cfrac { { p }^{ 2 } }{ { b }^{ 2 } } =\sin ^{ 2 }{ \alpha }$ , where

$k$ is equal to

0%
$\cfrac { 2pq }{ ab }$
0%
$-\cfrac { 2pq }{ ab }$
0%
$-\cfrac { pq }{ ab}$
0%
$\cfrac { pq }{ ab }$

Explanation

Given,

$\cos ^{ -1 }{ \left( \cfrac { p }{ a } \right) } +\cos ^{ -1 }{ \left( \cfrac { q }{ b } \right) } =\alpha \quad$

$\Rightarrow \cos ^{ -1 }{ \left[ \cfrac { p }{ a } .\cfrac { q }{ b } -\sqrt { \left( 1-\cfrac { { p }^{ 2 } }{ { a }^{ 2 } } \right) } \sqrt { \left( 1-\cfrac { { q }^{ 2 } }{ { b }^{ 2 } } \right) } \right] } =\alpha \quad \quad$

$\Rightarrow \cfrac { pq }{ ab } -\sqrt { \left( 1-\cfrac { { p }^{ 2 } }{ { a }^{ 2 } } \right) \left( 1-\cfrac { { q }^{ 2 } }{ { b }^{ 2 } } \right) } =\cos { \alpha }$

$\Rightarrow { \left( \cfrac { pq }{ ab } -\cos { \alpha } \right) }^{ 2 }=1-\cfrac { { p }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { q }^{ 2 } }{ { b }^{ 2 } } +\cfrac { { p }^{ 2 }{ q }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } =\cfrac { { p }^{ 2 }{ q }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } +\cos ^{ 2 }{ \alpha } -\cfrac { 2pq\cos { \alpha } }{ ab }$

$\Rightarrow \cfrac { { p }^{ 2 } }{ { a }^{ 2 } } -\cfrac { 2pq }{ ab } \cos { \alpha } +\cfrac { { q }^{ 2 } }{ { b }^{ 2 } } =1-\cos ^{ 2 }{ \alpha } =\sin ^{ 2 }{ \alpha }$
But

$\cfrac { { p }^{ 2 } }{ { a }^{ 2 } } +k\cos { \alpha } +\cfrac { { q }^{ 2 } }{ { b }^{ 2 } } =\sin ^{ 2 }{ \alpha }$

$\Rightarrow k=-\cfrac { 2pq }{ ab }$

The number of solutions of the equation

$\sin\theta +\cos\theta =\sin 2\theta$ in the interval

$[-\pi, \pi]$ is?

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

Explanation

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

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

$\therefore sin\theta+\cos\theta=\dfrac{1-\sqrt5}{2}$

$\Rightarrow\cos(\pi/4-\theta)=\dfrac{1-\sqrt5}{2}=-\dfrac{(\sqrt5-1)}{2}$

Thus, there are two possible solutions for

$\theta$ between

$-\pi$ to

$\pi$ .

The number of real solutions of the equation

$2\sin 3x+\sin 7x-3=0$ which lie in the interval

$[-2\pi, 2\pi]$ is?

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

Explanation

$2\text{ sin}3x+\text{ sin}7x=3$

For this to be true, both

$\text{sin}3x$ and

$\text{sin}7x$ must be equal to

$1$

Now,

For

$\text{ sin}3x=1$

$\Rightarrow x=\dfrac{\pi}{6},\dfrac{5\pi}{6},\dfrac{9\pi}{6},\dfrac{-\pi}{2},\dfrac{-7\pi}{6},\dfrac{-11\pi}{6}$

For

$\text{ sin}7x=1$

$\Rightarrow x=\dfrac{(4n+1)\pi}{14}$ for

$n=0,1,2,3,4,5,6$

$=-\dfrac{(4n-1)\pi}{14}$ for

$n=1,2,3,4,5,6,7$

So, only possible values of

$x$ are

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

$\dfrac{-\pi}{2}$

$\sin x =$

$\cos^2x$ , then

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

0%
$0$
0%
$1$
0%
$2$
0%
None of these

Explanation

Given,

$\sin x=\cos^2x$

Thus

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

$\Rightarrow \sin ^2x+\sin x=1$ ....(1)

We need to find value of

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

$\Rightarrow \sin x(1+\sin x)$

$\Rightarrow \sin x+\sin^2 x$

From the equation (1), we have

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

Option B is correct.

Find the value of

$\tan { 1 } \tan { 2 } ....\tan { 89 }$

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

Explanation

$\tan 1\cdot \tan 2.....\tan 89$

but

$\tan 89$ can be written as

$\cot 1[\tan (90-1)=\cot 1]$ by

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

likewise...,

$\tan 88=\cot 2$

$\tan 87=\cot 3$

....... up to

$\tan 46=\cot 44$

then middle one is

$\tan 45=1$

so it is

$\tan 1\cdot \tan 2....\tan 44\cdot \tan 45\cdot \cot 44.....\cot 1$

$\tan$ and

$\cot$ cancel out by

$[\tan(\theta)^{\ast}\cot(\theta)=1]$

so, remaining is

$1$

answer

$=1$ .

1150365_699611_ans_8caf925322ed44beb35bcf6ad0aa2347.jpg

The value of

$\cos ^{ 4 }{ \left( \cfrac { \pi }{ 8 } \right) } +\cos ^{ 4 }{ \left( \cfrac { 3\pi }{ 8 } \right) } +\cos ^{ 4 }{ \left( \cfrac { 5\pi }{ 8 } \right) } +\cos ^{ 4 }{ \left( \cfrac { 7\pi }{ 8 } \right) }$ is

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

Explanation

Since

$\boxed{\cfrac { 1+\cos 2{ \theta } }{ 2 } =\cos ^{ 2 }{ \theta }}$

$\Rightarrow \cos^4(\dfrac{\pi}{8})={ \left[ \cfrac { 1+\cos { (2\pi /8) } }{ 2 } \right] }^{ 2 }={ \left[ \cfrac { 1+\cos { (\pi /4) } }{ 2 } \right] }^{ 2 }$
Similarly

$\cos ^{ 4 }{ \cfrac { 3\pi }{ 8 } } ={ \left[ \cfrac { 1+\cos { (3\pi /4) } }{ 2 } \right] }^{ 2 }$

$\cos ^{ 4 }{ \cfrac { 5\pi }{ 8 } } ={ \left[ \cfrac { 1+\cos { (5\pi /4) } }{ 2 } \right] }^{ 2 }$

$\cos ^{ 4 }{ \cfrac { 7\pi }{ 8 } } ={ \left[ \cfrac { 1+\cos { (7\pi /4) } }{ 2 } \right] }^{ 2 }$

$\therefore$ Given expression will become

$\cfrac { 1 }{ 4 } \left[ 2{ \left( 1+\cfrac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }+2{ \left( 1+\cfrac { 1 }{ \sqrt { 2 } } \right) }^{ 2 } \right] =\cfrac { 1 }{ 4 } \times 2(2+1)=\cfrac { 1 }{ 4 } \times 2\times 3=\cfrac { 3 }{ 2 }$

Let

$X = \left \{x\epsilon \mathbb {R} : \cos (\sin x) = \sin (\cos x)\right \}$ . The number of solutions of

$X$ is?

0%
$0$
0%
$2$
0%
$4$
0%
Not finite

Explanation

$\cos (\sin x)=\sin (\cos x)\implies \cos (\sin x)=\cos (\dfrac{\pi}{2}-\cos x)$

$\sin x=\dfrac{\pi}{2}-\cos x\implies \sin x+\cos x=\dfrac{\pi}{2}=1.72$

But

$\sin x+\cos x$ maximum value is

$\sqrt{2}=1.414$

$\sin x+\cos x\neq \dfrac{\pi}{2}$

Number of solutions of

$X$ is

$0$

The principal solution of

$\sec x = \dfrac {2}{\sqrt {3}}$ are

0%
$\dfrac {\pi}{3}, \dfrac {11\pi}{6}$
0%
$\dfrac {\pi}{6}, \dfrac {11\pi}{6}$
0%
$\dfrac {\pi}{4}, \dfrac {11\pi}{4}$
0%
$\dfrac {\pi}{3}, \dfrac {11\pi}{4}$

Explanation

Given

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

We know that,

$\sec\: \dfrac{\pi}{6}=\dfrac{2}{\sqrt{3}}$ also

$\sec \left(2\pi-\dfrac{\pi}{6}\right)=\sec\: \dfrac{11\pi}{6}=\dfrac{2}{\sqrt{3}}$

Therefore the principal solutions are

$\dfrac{\pi}{6}, \dfrac{11\pi}{6}$

The equation

$2\tan { x } +5x-2=0$ has:

0%
no solution in $\left[ 0,\dfrac{\pi}{4} \right]$
0%
at least one real solution in $\left[ 0,\dfrac{\pi}{4} \right]$
0%
two real solution in $\left[ 0,\dfrac{\pi}{4} \right]$
0%
None of these

Explanation

Given that

$2 \tan x + 5x -2 = 0$

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

The functions

$y = \tan x$ and

$y = \dfrac{-5}{2}x + 1$ can be plotted on the same graph.

The solution to the equation

$\tan x = \dfrac{-5}{2}x + 1$ will be at the intersection of the two curves.

The graph is given in the figure above.

From the graph, it is seen that the line intersects the x-axis at

$x = \dfrac{2}{5}$ .

For some

$x$ such that

$0 < x <\dfrac{2}{5}$ , the 2 curves meet.

Since

$0 <\dfrac{2}{5} < \dfrac{\pi}{4}$ ,

Hence, B is true, i.e there is at least one real solution in

$\left[0, \dfrac{\pi}{4}\right]$ .

The expression

${ \left( \tan { \theta } +\sec { \theta } \right) }^{ 2 }$ is equal to

0%
$\cfrac { 1+\cos { \theta } }{ 1-\cos { \theta } }$
0%
$\cfrac { 1+\sin { \theta } }{ 1-\sin { \theta } }$
0%
$\cfrac { 1-\cos { \theta } }{ 1+\cos { \theta } }$
0%
$\cfrac { 1-\sin {\theta } }{ 1+\sin { \theta } }$

Explanation

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

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

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

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

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

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

$\dfrac {\tan \theta}{1 +\tan^{2} \theta} + \dfrac {\cot \theta}{(1 + \cot^{2}\theta)^{2}}$ is equal to

0%
$2\sin \theta \cdot \cos \theta$
0%
$\text{cosec} \theta \cdot \sec \theta$
0%
$\sin \theta \cdot \cos \theta$
0%
$2\text{cosec} \theta \cdot \sec \theta$

Explanation

$\dfrac {\tan \theta}{(1 + \tan^{2}\theta)^{2}} + \dfrac {\cot \theta}{(1 + \cot^{2}\theta)^{2}} = \dfrac {\tan \theta}{\sec^{4}\theta} + \dfrac {\cot \theta}{\text{cosec}^{4}\theta}$

$= \dfrac {\text{cosec} \theta \sec \theta \left (\dfrac {1}{\sin^{2}\theta} + \dfrac {1}{\cos^{2}\theta}\right )}{\sec^{4}\theta \text{cosec}^{4}\theta} = \dfrac {\sec^{2}\theta \text{cosec}^{2}\theta}{\sec^{3} \theta \text{cosec}^{3}\theta} = \sin \theta \cos \theta$ .

$\sin \alpha + \cos \alpha = k$ , then

$|\sin \alpha - \cos \alpha |$ equals.

0%
$\sqrt {2 - k^{2}}$
0%
$\sqrt {k^{2} - 2}$
0%
$|k|$
0%
$\sqrt {2} - k$
0%
$k - \sqrt {2}$

Explanation

$\sin\alpha +\cos\alpha =k$

${ \left( \sin\alpha +\cos\alpha \right) }^{ 2 }={ k }^{ 2 }$

${ \sin }^{ 2 }\alpha +{ \cos }^{ 2 }\alpha +2\sin\alpha \cos\alpha ={ k }^{ 2 }$

$2\sin\alpha \cos\alpha ={ k }^{ 2 }-1$

${ \left| \sin\alpha -\cos\alpha \right| }^{ 2 }=\left| { \sin }^{ 2 }\alpha +{ \cos }^{ 2 }\alpha -2\sin\alpha \cos\alpha \right|$

${ \left| \sin\alpha -\cos\alpha \right| }^{ 2 }=\left| 1-\left( { k }^{ 2 }-1 \right) \right|$

${ \left| \sin\alpha -\cos\alpha \right| }^{ 2 }=\left| 2-{ k }^{ 2 } \right|$

$\left| \sin\alpha -\cos\alpha \right| =\sqrt { 2-{ k }^{ 2 } }$

$n = \dfrac {\cos \alpha}{\cos \beta}, m = \dfrac {\sin \alpha}{\sin \beta}$ , then

$(m^{2} - n^{2})\sin^{2}\beta$ is

0%
$1 - n$
0%
$1 + n$
0%
$1 - n^{2}$
0%
$1 + n^{2}$

Explanation

Given,

$(m^2-n^2)\sin ^2\beta$

$=\left ( \dfrac{\sin ^2\alpha }{\sin ^2\beta } -\dfrac{\cos ^2\alpha }{\cos ^2\beta }\right )\sin ^2\beta$

$=\left ( \dfrac{\cos ^2\beta \sin ^2\alpha-\sin ^2\beta\cos ^2\alpha }{\sin ^2\beta \cos ^2\beta} \right )\sin ^2\beta$

$=\sin ^2\alpha-\dfrac{\sin ^2\beta}{\cos ^2\beta }\cos ^2\alpha$

$=1-\dfrac{\cos ^2\alpha }{\cos ^2\beta }$

$=1-n^2$

$\sin { \theta } =\cfrac { 8 }{ 17 }$ where

${ 0 }^{ o }<\theta <{ 90 }^{ o }$ , then

$\tan { \theta } +\sec { \theta }$ is

0%
$\dfrac {1}{3}$
0%
$\dfrac {2}{3}$
0%
$\dfrac {4}{3}$
0%
$\dfrac {5}{3}$

Explanation

$\sin\theta =8/17,\quad \tan\theta =\dfrac { sin\theta }{ \sqrt { 1-{ sin }^{ 2 }\theta } } =8/15$

$$\sec\theta =\dfrac { 1 }{ \cos\theta } =\dfrac { 1 }{ \sqrt { 1-{

\sin }^{ 2 }\theta } } =\dfrac { 17 }{ 15 } $$

$\tan\theta +\sec\theta =\dfrac { 8 }{ 15 } +\dfrac { 17 }{ 15 } =\dfrac { 25 }{ 15 } =\dfrac { 5 }{ 3 }$

$\cfrac { 3-4\sin ^{ 2 }{ \theta } }{ \cos ^{ 2 }{ \theta } }$ is equal to

0%
$3-\cot ^{ 2 }{ \theta }$
0%
$3+\cot ^{ 2 }{ \theta }$
0%
$3-\tan ^{ 2 }{ \theta }$
0%
$3+\tan ^{ 2 }{ \theta }$

Explanation

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

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

$\sec^26 + \text{cosec}^26$ is equal to:

0%
$\sec^26. \cot^26$
0%
$\sec^26 .\tan^26$
0%
$\text{cosec}^26. \cot^26$
0%
$\text{cosec}^26. \sec^26$

Explanation

${ \sec }^{ 2 }6+ \text{ cosec}^{ 2 }6$

$= \dfrac { 1 }{ { \cos }^{ 2 }6 } +\dfrac { 1 }{ { \sin }^{ 2 }6 }$

$= \dfrac { { \sin }^{ 2 }6+{ \cos }^{ 2 }6 }{ { \sin }^{ 2 }6{ \cos }^{ 2 }6 }$

$={ \sec }^{ 2 }6 \text{ cosec}^{ 2 }6$

$\sin { \theta } +\cos { \theta } =\sqrt { 2 }$ and

$\theta$ is acute, then

$\tan{\theta}$ is

0%
$\dfrac {1}{\sqrt { 3 } }$
0%
$1$
0%
$\sqrt { 3 }$
0%
$\infty$

Explanation

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

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

$\Rightarrow \sin { 2\theta } =1\Rightarrow \cfrac { 2\tan ^{ 2 }{ \theta } }{ 1+\tan ^{ 2 }{ \theta } } =1\Rightarrow \tan ^{ 2 }{ \theta }=1 \Rightarrow \tan { \theta =\pm 1\quad }$
since

$\theta$ is acute

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

$\tan { \theta } =\dfrac {3}{4}$ and

$0<\theta <{ 90 }^{ 0 }$ , then the value of

$\sin { \theta } \cos { \theta }$ is

0%
$\dfrac {1}{5}$
0%
$\dfrac {9}{5}$
0%
$\dfrac {12}{25}$
0%
$\dfrac {25}{12}$

Explanation

$\tan { \theta } =\cfrac { 3 }{ 4 } \Rightarrow \cfrac { P }{ B } =\cfrac { 3 }{ 4 } \Rightarrow H=\sqrt { 9+16 } =5\Rightarrow \sin { \theta } \cos { \theta } =\cfrac { 12 }{ 25 }$

$\left( \text{cosec} { \theta } -\sin { \theta } \right) \left( \sec { \theta } -\cos { \theta } \right) \left( \tan { \theta } +\cot { \theta } \right)$ simplifies to

0%
$0$
0%
$1$
0%
$\tan { \theta }$
0%
$\cot { \theta }$

Explanation

$\left( \cos\theta -\sin\theta \right) \left( \sec\theta -\cos\theta \right) \left( \tan\theta +\cot\theta \right)$

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

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

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

$=1$

The expression

$3(\sin x - \cos x)^{4} + 6(\sin x + \cos x)^{2} + 4(\sin^{6}x + \cos^{6}c)$ is equal to

0%
$10$
0%
$11$
0%
$12$
0%
$13$

Explanation

$3(\sin x - \cos x)^{4} + 6(\sin x + \cos x)^{2} + 4(\sin^{6}x + \cos^{6}c)$

$= 3(1 + 4\sin^{2} x\cos^{2} x - 4\sin x\cos x) + 6(1 + 2\sin x \cos x) + 4(1 - 3\sin^{2}x \cos^{2}x) = 3 + 6 + 4 = 13$ .

$(\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta)$ can be written as

0%
$\sin \theta + \cos \theta$
0%
$\sin^{3} \theta - \cos^{3}\theta$
0%
$\sin^{3}\theta + \cos^{3}\theta$
0%
$\sin \theta - \cos \theta$

Explanation

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

$= \cos \theta (1 - \sin^{2} \theta) + \sin \theta (1 - \cos^{2}\theta) = \cos^{3}\theta + \sin^{3}\theta$ .

$\sin ^{ 6 }{ \theta } +\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } -\sin ^{ 4 }{ \theta } \cos ^{ 4 }{ \theta } -\cos ^{ 6 }{ \theta }$ equals to

0%
$\sin ^{ 2 }{ \theta } -\cos ^{ 3 }{ \theta }$
0%
$\sin ^{ 3 }{ \theta } -\cos ^{ 3 }{ \theta } \quad$
0%
$\sin ^{ 4 }{ \theta } +\cos ^{ 4 }{ \theta }$
0%
$\sin ^{ 2 }{ \theta } -\cos ^{ 2 }{ \theta }$

$\left( 1-\sin ^{ 2 }{ \theta } \right) \left( 1+\tan ^{ 2 }{ \theta } \right)$ is equal to

0%
$1$
0%
$1.5$
0%
$2$
0%
$2.5$

Explanation

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

$\tan { \theta } \left( 1-\cot ^{ 2 }{ \theta } \right)$ is equal to

0%
$\cot { \theta } \left( 1-\tan ^{ 2 }{ \theta } \right)$
0%
$\cot { \theta } \left( \tan ^{ 2 }{ \theta } -1 \right)$
0%
$\cot { \theta } \tan ^{ 2 }{ \theta }$
0%
$\tan { \theta } co\sec ^{ 2 }{ \theta }$

Explanation

$\rightarrow \tan\theta \left( 1-{ cot }^{ 2 }\theta \right)$

$\rightarrow \dfrac { 1 }{ \cot\theta } \left( 1-\dfrac { 1 }{ { \tan }^{ 2 }\theta } \right)$

$\rightarrow \dfrac { 1 }{ \cot\theta .{ \tan }^{ 2 }\theta } \left( { \tan }^{ 2 }\theta -1 \right)$

$\rightarrow \cot\theta \left( { \tan }^{ 2 }\theta -1 \right)$

$2\cos A+3\cos B+5\cos C$

$=2\sin A+3\sin B+5\sin C=0$ then

$8\cos 3A+27\cos 3B+125\cos C$

$=k\cos(A+B+C)$ then

$k=$

0%
$70$
0%
$80$
0%
$90$
0%
$60$

Explanation

$\Rightarrow$

$2\cos { A } +3\cos { B } +5\cos { C } =2\sin { A } +3\sin { B } +5\sin { C } =0$

$\Rightarrow$

$\cos { \theta } =\cfrac { 1 }{ 2 } ({ e }^{ j\theta }+{ e }^{ -j\theta })$

$\Rightarrow$

$\sin { \theta } =\cfrac { 1 }{ 2j } ({ e }^{ j\theta }-{ e }^{ -j\theta })$

$\therefore 2\cfrac { 1 }{ 2 } ({ e }^{ jA }+{ e }^{ -jA })+3\cfrac { 1 }{ 2 } ({ e }^{ jB }+{ e }^{ -jB })+5\cfrac { 1 }{ 2 } ({ e }^{ jC }+{ e }^{ -jC })=0$

$\Rightarrow$

$2({ e }^{ jA }+{ e }^{ -jA })+3({ e }^{ jB }+{ e }^{ -jB })+5({ e }^{ jC }+{ e }^{ -jC })=0\quad -(i)$

Also,

$\\ 2({ e }^{ jA }-{ e }^{ -jA })+3({ e }^{ jB }-{ e }^{ -jB })+5({ e }^{ jC }-{ e }^{ -jC })=0\quad -(ii)$

$\Rightarrow$ As

$(i)-(ii)$

$\Rightarrow$

$4{ e }^{ -jA }+6{ e }^{ -jB }+10{ e }^{ -jC }=0$

$\Rightarrow$

$2{ e }^{ -jA }+3{ e }^{ -jB }+5{ e }^{ -jC }=0$

Cubing on both sides.

$\Rightarrow$

$8{ e }^{ -j3A }+27{ e }^{ -j3B }+125{ e }^{ -j3C }+3({ e }^{ -jA }\times 2\times 3{ e }^{ -jB }.5{ e }^{ -jC })=0$

$\therefore 8{ e }^{ -j3A }+27{ e }^{ -j3B }+125{ e }^{ -j3C }=90{ e }^{ -jA }{ e }^{ -jB }{ e }^{ -jC }$

$\therefore k=90$

$\sin \left(\sin^{-1}\dfrac{1}{5}+\cos ^{-1}x\right)=1$ , then find the value of

$x$ .

0%
$-\dfrac{1}{5}$
0%
$\dfrac{1}{2}$
0%
$-\dfrac{1}{2}$
0%
$\dfrac{1}{5}$

Explanation

We have,

$\sin \left( {{\sin }^{-1}}\dfrac{1}{5}+{{\cos }^{-1}}x \right)=1$

${{\sin }^{-1}}\dfrac{1}{5}+{{\cos }^{-1}}x={{\sin }^{-1}}1$

${{\sin }^{-1}}\dfrac{1}{5}+{{\cos }^{-1}}x=\dfrac{\pi }{2}$

${{\sin }^{-1}}\dfrac{1}{5}=\dfrac{\pi }{2}-{{\cos }^{-1}}x$

${{\sin }^{-1}}\dfrac{1}{5}={{\sin }^{-1}}x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore \left( {{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2} \right)$

$x=\dfrac{1}{5}$

Hence, this is the answer.

Solve:

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

0%
$\dfrac{1+\cos \theta}{\sin\theta}$
0%
$\dfrac{\sin\theta}{1-\cos\theta}$
0%
$\dfrac{1-\cos\theta}{\sin\theta}$
0%
$\dfrac{\cos\theta}{1-\sin\theta}$

Explanation

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

$=\dfrac{\cot \theta+\text{cosec }\theta-( \text{cosec}^2 \theta-\cot^2\, \theta)}{\cot \theta-\text{cosec }\theta+1}$

$=\dfrac{(\cot \theta+\text{cosec }\theta)[1+\cot \theta-\text{cosec }\theta]}{1+\cot \theta-\text{cosec }\theta}$

$=\cot \theta+\text{cosec }\theta$

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

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

$\cos { \theta } -\sin { \theta } =\sqrt { 2 } \sin { \theta }$ , then

$\cos { \theta } +\sin { \theta }$ is

0%
$\sqrt { 2 } \cos { \theta }$
0%
$\sqrt { 2 } \sin { \theta }$
0%
$0$
0%
$1$

Explanation

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

$\rightarrow { \left( \sin \theta +\cos \theta \right) }^{ 2 }+2{ \sin }^{ 2 }\theta =2\Rightarrow \sin \theta +\cos \theta =\sqrt { 2-2{ \sin }^{ 2 }\theta }$

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

$\theta$ and

$\phi$ are angles in the first quadrant such that

$\tan { \theta } =\dfrac 17$ and

$\sin { \phi } =\dfrac {1}{\sqrt { 10 }}$ , then

0%
$\theta +2\phi ={{90 }}^{o}$
0%
$\theta +2\phi ={{ 30 }}^{o}$
0%
$\theta +2\phi ={{ 75 }}^{o}$
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$\theta +2\phi ={{45}}^{o}$

Explanation

Since

$\tan { \theta } =\dfrac 17$ we have

$\sin { \theta } =\dfrac 1{\left( 5\sqrt { 2 } \right)}$ and

$\cos { \theta } =\dfrac{7}{\left( 5\sqrt { 2 } \right)}$
Also

$\sin { \phi } =\dfrac {1}{\sqrt { 10 }} \Rightarrow \tan { \phi } =\dfrac{1}{3}$

$\therefore\quad \sin { 2\phi } =\dfrac{\left( 2\tan { \phi } \right)} {\left( 1+\tan ^{ 2 }{ \phi } \right)}$

$=\dfrac {\left( \dfrac 23 \right)} {\left[ 1+\left( \dfrac 19 \right) \right]} =\dfrac 35$
and so

$\cos { 2\phi } =\dfrac 45$
Hence

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

$=\cfrac { 1 }{ 5\surd \left( 2 \right) } .\cfrac { 4 }{ 5 } +\cfrac { 7 }{ 5\surd \left( 2 \right) } .\cfrac { 3 }{ 5 } =\cfrac { 25 }{ 25\surd \left( 2 \right) } =\cfrac { 1 }{ \surd \left( 2 \right) }$

$\theta +2\phi ={45}^{o}$

The value of

$\cos { \left( \pi /5 \right) } \cos { \left( 2\pi /5 \right) } \cos { \left( 4\pi /5 \right) } \cos { \left( 8\pi /5 \right) }$ is

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$\dfrac{1}{16}$
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$0$
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$\dfrac{-1}{8}$
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$\dfrac{-1}{16}$

Explanation

$\cos { \left( \pi /5 \right) } \cos { \left( 2\pi /5 \right) } \cos { \left( 4\pi /5 \right) } \cos { \left( 8\pi /5 \right) } =\cfrac { 1 }{ { 2 }^{ 4 }\sin { \left( \pi /5 \right) } } \sin { \cfrac { { 2 }^{ 4 }\pi }{ 5 } }$

$=-\cfrac { 1 }{ 16 } \left[ \because\quad \sin { \cfrac { 16\pi }{ 5 } =\sin { \left( 3\pi +\cfrac { \pi }{ 5 } \right) } =-\sin { \cfrac { \pi }{ 5 } } } \right]$

CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 7 - 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 7 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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