MCQ Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 3

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

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

Explanation

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

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

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

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

$=1$

$\dfrac{\cos A}{1-\tan A} + \dfrac{\sin A}{1 - \cot A}$ is equal to

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$\sin A - \cos A$
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$\sin A + \cos A$
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$1 -\cos A$
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$1 - \sin A$

Explanation

Simplifying the above equation we get

$\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-cotA}$

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

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

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

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

$=\cos A+\sin A$

Hence answer is B

$(\cos A + \sin A)^2 - (\cos A - \sin A)^2$ is equal to

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$-1$
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$2$
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$0$
0%
None of the above

Explanation

Expanding the given expression we get

$(\cos A+\sin A)\:^2-(\cos A+\sin A)\:^2$

$=\cos A\:^2+\sin A\:^2+2\cos A \sin A-(\cos A\:^2+\sin A\:^2-2\cos A \sin A)$

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

$=4 \cos A \sin A$

Hence, option D is correct.

$\sin$

$\beta=\dfrac{12}{13},$ then the value of

$\dfrac{13\sin\beta+5 \sec\beta}{5\tan\beta+6 \csc\beta}$

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$0$
0%
$1$
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$\dfrac{50}{37}$
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$\dfrac{37}{2}$

Explanation

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

$\cos\beta=\sqrt{1-sin^2\beta}=\dfrac{5}{13}$

Hence

$\tan\beta=\dfrac{12}{5}$

Therefore

$\dfrac{13\sin\beta+5\sec\beta}{5\tan\beta+6\csc\beta}$

$=\dfrac{12+13}{12+\dfrac{13}{2}}$

$=\dfrac{50}{37}$

Hence answer is C

Th value of

$(\sin \theta + \csc \theta)^2 + (\cos\theta + \sec \theta)^2 - (\tan^2 \theta+\cot^2 \theta)$ is

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$1$
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$2$
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$6$
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$7$

Explanation

By expanding the expression we get

$(\sin\theta+\csc\theta)\:^2+(\cos\theta+sec\theta)\:^2-(\tan^2\theta+cot^2\theta)$

$=\sin^2\theta+\csc^2\theta+2+sec^2\theta+\cos^2\theta+2-\tan^2\theta-cot^2\theta$

$=4+(\sin^2\theta+\cos^2\theta)+(\csc^2\theta-cot^2\theta)+(sec^2\theta-\tan^2\theta)$

$=4+1+1+1=7$

Hence answer is D

$\tan \alpha=2\sqrt{2},$ then the value of

$\dfrac{\tan\alpha}{\dfrac{\sin^{3}\alpha}{\cos\alpha}+\sin\alpha.\cos\alpha}$ is

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

Explanation

Simplifying the given expression we get

$\dfrac{\dfrac{\sin\alpha}{\cos\alpha}}{\sin^2\alpha\:\tan\alpha+\sin\alpha \cos\alpha}$

$=\dfrac{\sin\alpha}{(\sin\alpha\:\cos\alpha)(\sin\alpha\:\tan\alpha+\cos\alpha)}$ ...(taking $\sin\alpha$ common from the denominator)

$=\dfrac{1}{\sin\alpha\:\tan\alpha\:\cos\alpha+\cos^2\alpha}$

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

$=1$

Hence answer is D

$(1-\sin^2 \theta)(1+\tan^2 \theta)$ is equal to

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$1$
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$1.5$
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$2$
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$2.5$

Explanation

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

$\cos A$

$=$ 2 sin A, find the value of

$cosec A$ .

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$1$
0%
$_{-}^{+}\sqrt{5}$
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$\dfrac{1}{2}$
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$_{-}^{+}\sqrt{3}$

Explanation

$\cos A=2\sin A$

Dividing by $\sin A$ on both the sides we get

$\cot A=2$ ...(i)

$cosec^2A=1+\cot^2A$

$cosec^2A=1+2^2$ ...(from i )

$cosec^2A=5$

Therefore

$cosec A=\pm\sqrt{5}$

Hence answer is B

The value of

$\sin^6 \theta+\cos^6 \theta+3 \sin^2\theta.\cos^2\theta$ is

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

Explanation

$(\sin^2\theta+\cos^2\theta)\:^3=1$

$\sin^6\theta+\cos^6\theta+3\sin^2\theta \cos^2\theta(\sin^2\theta+\cos^2\theta)=1$

$\sin^6\theta+\cos^6\theta+3\sin^2\theta \cos^2\theta(1)=1$

Hence answer is C

$x, y \in [0,2\pi]$ , then total number of ordered pairs

$(x, y)$ satisfying the equation

$\sin x.\cos y=1$ is equal to

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$1$
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$3$
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$5$
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$7$

Explanation

$\sin x. \cos y =1$

$\Rightarrow \sin x=1 , \cos y =1$ or

$\sin x =- 1 , \cos y = -1$
Hence, the ordered pairs of

$(x,y)$ are

$(\displaystyle \frac {\pi}{2} , 0)$ ,

$(\displaystyle \frac {\pi}{2} , 2\pi)$ ,

$( \displaystyle \frac {3\pi}{2} , \pi)$

The set of angles between

$0$ &

$2\pi$ satisfying the equation

$4 \cos^2\theta -2\sqrt 2 \cos\theta-1=0$ is-

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$\left \{\dfrac {\pi}{12}, \dfrac {5\pi}{12}, \dfrac {19\pi}{12}, \dfrac {23\pi}{12}\right \}$
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$\left \{\dfrac {\pi}{12}, \dfrac {7\pi}{12}, \dfrac {17\pi}{12}, \dfrac {23\pi}{12}\right \}$
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$\left \{\dfrac {5\pi}{12}, \dfrac {13\pi}{12}, \dfrac {19\pi}{12}\right \}$
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$\left \{\dfrac {\pi}{12}, \dfrac {7\pi}{12}, \dfrac {19\pi}{12}, \dfrac {23\pi}{12}\right \}$

Explanation

$\cos \theta=\dfrac {2\sqrt 2\pm \sqrt {8+16}}{8}=\dfrac {\sqrt 2\pm \sqrt 6}{4}$
$\cos \theta =\dfrac {\sqrt 6+\sqrt 2}{4}$
$\Rightarrow \theta =\dfrac {\pi}{12}; 2\pi -\dfrac {\pi}{12}=\dfrac {23\pi}{12}$
$\cos \theta=-\dfrac {\sqrt 6-\sqrt 2}{4}$
$\cos \theta =\cos (\pi -\dfrac{5\pi}{12}); \cos (\pi +\dfrac{5\pi}{12})$
$\theta=\dfrac{7\pi}{12}; \dfrac{17\pi }{12}$
Hence option B is the correct option

Number of values of

$\theta \epsilon [0, 2\pi]$ satisfying the equation

$\cot x-\cos x=1-\cot x.\cos x$

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1
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2
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3
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4

Explanation

$\dfrac{cosx}{sinx}-cosx = 1 - \dfrac{cos^2x}{sinx}$

$\Rightarrow cosx-cosx*sinx=sinx-cos^2x$

$\Rightarrow cos^2x+cosx-cosxsinx-sinx=0$

$\Rightarrow cosx(1+cosx)-sinx(1+cosx)=0$

$\Rightarrow (cosx-sinx)(1+cosx)=0$

$\Rightarrow x = \dfrac{\pi}{4} , \dfrac{5\pi}{4} , {\pi}$

Hence, 3 solutions exist.

$\alpha + \beta =90^0$ and

$\alpha = 2\beta,$ then

$\cos^2 \alpha + \sin^2 \beta$ equals to

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

Explanation

$\alpha+\beta=90^0$ ...(i)

$\alpha-2\beta=0$ ...(ii)

Subtracting ii from i we get

$3\beta=90^0$

$\beta=30^0$

Therefore

$\cos^2\alpha+\sin ^2\beta$

$=\cos^2(90^0-\beta)+\sin ^2\beta$ ...from(i)

$=\sin ^2\beta+\sin ^2\beta$

$=2\sin ^2\beta$

Substituting the value of $\beta$ we get

$2\sin ^2\beta=2(\dfrac{1}{2})\:^2$

$=\dfrac{1}{2}$

Hence the answer is A

Which of the following relations is/are an identity?

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$\displaystyle \sin x^2=2 \sin\displaystyle \dfrac {x^2}{2} \cos \displaystyle \dfrac {x^2}{2}$
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$\displaystyle \sin x=\dfrac {2 \tan\displaystyle \dfrac {x}{2}}{1+\tan^2\displaystyle \dfrac {x}{2}}$
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$\log xy=\log |x| + \log |y|$
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$\sqrt {1-\sin^2x}=\cos x$

Explanation

Option

$A$ is an identity as

$\sin2x=2\sin x\cos x\Rightarrow \sin{ x }^{ 2 }=2\sin\dfrac { { x }^{ 2 } }{ 2 } \cos\dfrac { { x }^{ 2 } }{ 2 }$
Option

$B$ is also Identity
Option

$C$ is wrong as

$\log xy$ is define only when

$x>0$ and

$y>0$ or

$x<0$ and

$y<0$ . However, if

$xy<0$ , RHS will still be defined, hence not an identity.
Option

$D$ is wrong as

$\sqrt { 1-{ \sin }^{ 2 }x } =\pm \cos x$
Hence, options

$A$ and

$B$ are correct.

Total number of solution of

$\sin x=x+\dfrac {1}{x} x\epsilon (0, 2\pi)$ , is equal to-

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

Explanation

As we know

$AM >= GM$

$\sin x=x+\dfrac {1}{x}, x\epsilon (0, 2\pi)$ R. H. S,

$x+\dfrac {1}{x}\geq 2$
While

$\sin x\leq 1$
Option B

The number of solutions of the equation

$x^3+2x^2+5x+2 cos x=0$ in

$[0, 2\pi]$ is:

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0
0%
1
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2
0%
3

Explanation

Let

$f(x)=x^3+2x^2+5x+2 \cos x$

$\Rightarrow f'(x)=3x^2+4x+5-2 sinx=2(x+\frac {2}{3})^2+\frac {11}{3}-2 sin x$

Now

$\frac {11}{3}-2 sinx > 0$

$x(as -1 \leq sin x \leq 1)$

$\Rightarrow f'(x) > 0$

$x\Rightarrow f(x)$ is an increasing function.

Now

$f(0)=2$

$\Rightarrow f(x)=0$ has no solution in

$[0, 2\pi]$ .

Hence (A) is the correct answer.

For each integer

$n>1$ , let

$S(n)$ denote the number of solution of the equation

$\sin x=\sin nx$ on the interval

$[0,\pi]$ Find the value of

$S(2)+S(3)+S(4)$ .

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12
0%
11
0%
15
0%
10

Explanation

$sin(x)=sin2x$

$sin(x)=2sin(x).cos(x)$

$sin(x)[2cos(x)-1]=0$

$sin(x)=0$ and

$cos(x)=\dfrac{1}{2}$

$x=0,\pi$ and

$x=\dfrac{\pi}{3}$ ...(i)
Hence there are 3 solutions.

$S(3)$

$sin(x)=sin3x$

$sin3x-sinx=0$

$2cos(2x).sinx=0$

$sin(x)=0$ and

$cos(2x)=0$

$x=0,\pi$ and

$x=\dfrac{\pi}{4},\dfrac{3\pi}{4}$
Hence number of solutions are 4.

$S(4)$

$sin(x)=sin4x$

$sin4x-sinx=0$

$2cos(\dfrac{5x}{2}).sin(\dfrac{3x}{2})=0$

$x=0,\dfrac{2\pi}{3}$ and

$x=\dfrac{\pi}{5},\dfrac{3\pi}{5},\pi$
Hence number of solutions are 5.
Thus

$S(2)+S(3)+S(4)$

$=3+4+5$

$=12$ .

$0\leq x\leq 2\pi$ ,

$0\leq y\leq 2\pi$ and

$\sin x+\sin y=2$ then the value of

$x+y$ is

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$\pi$
0%
$\dfrac{\pi }{2}$
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$3\pi$
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none of these

Explanation

We have,

$\sin x + \sin y = 2$

but we know that maximum value of \sin x is

$1$

hence only solution is

$\sin x = 1$ and

$\sin y = 1$

so in

$0 \leq x \leq 2\pi, 0 \leq y \leq 2\pi$

solution is

$x = \dfrac{\pi}{2}, y = \dfrac{\pi}{2}$

therefore,

$x + y = \pi$

Hence, option 'A' is correct.

The smallest positive integral value of p for which the equation

$\cos \left ( p\sin x \right )=\sin \left ( p\cos x \right )$ in x has a solution in

$\left [ 0, 2\pi \right ]$ is

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2
0%
1
0%
3
0%
none of these

Explanation

$\cos(p \sin x) = \sin(p \cos x)$

$\Rightarrow \cos( p \sin x) = \cos\left(\dfrac{\pi}{2} - p \cos x \right)$

$\Rightarrow p \sin x = \dfrac{\pi}{2} - p \cos x \Rightarrow \cos x + \sin x = \dfrac{\pi}{2p}$

condition for this equation to have solution is,

$\dfrac{\pi}{2p} \leq \sqrt{2} \Rightarrow p \geq\dfrac{\pi}{2\sqrt{2}}$

Hence smallest positive integer is

$2$ .

The number of solutions of equation

$5\sec \theta-13=12\tan \theta$ in

$\left [ 0, 2\pi \right ]$ is

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

Explanation

$5 sec\theta - 13 = 12 tan\theta$

$\Rightarrow 12 sin\theta + 13 cos\theta = 5$

$\Rightarrow \dfrac{12}{\sqrt{313}} sin\theta + \dfrac{13}{\sqrt{313}} cos\theta = \dfrac{5}{\sqrt{313}}$

Let

$cos\alpha = \dfrac{13}{\sqrt{313}} \Rightarrow sin\alpha = \dfrac{12}{\sqrt{313}}$

$\Rightarrow cos\theta.cos\alpha + sin\theta.sin\alpha = \dfrac{5}{\sqrt{313}}$

$\Rightarrow cos(\theta - \alpha) = \dfrac{5}{\sqrt{313}}$

Hence solution in the given interval is,

$\theta = \alpha - cos^{-1}(\dfrac{5}{\sqrt{313}}), \alpha + cos^{-1}(\dfrac{5}{\sqrt{313}})$

Hence, option 'A' is correct.

$3\sin ^{2}\theta +2\sin ^{2}\phi=0$ and

$3\sin 2\theta +2\sin 2\phi=0$ ,

$0< \theta < \frac{\pi }{2}$ and

$0< \phi < \frac{\pi }{2}$ , then the value of

$\theta +2\phi$

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$\dfrac{\pi }{2}$
0%
$\dfrac{\pi }{4}$
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$0$
0%
none of these

$\sin \theta = a$ for exactly one value of

$\displaystyle \:\theta \in \left [ 0,\frac{7\pi }{3} \right ]$ then the value of a is

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$\displaystyle \:\frac{\sqrt{3}}{2}$
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1
0%
0
0%
-1

Explanation

We have,

$\theta \in [0,\dfrac{7\pi}{3}]$

$=[0,2\pi +\dfrac{\pi}{3}]$

$=[0,2\pi]+[2\pi,\dfrac{7\pi}{3}]$

$=T+\left[2\pi,\dfrac{7\pi}{3} \right]$ Where T stands for time period of the sinx.

Now

$sin\theta$ achieves a minimum value of -1 once in its time period, and maximum value of 1, only once in its time period.

Now the given interval is

$T+[2\pi,\dfrac{7\pi}{3}]$

In the interval of

$[2\pi,\dfrac{7\pi}{3}]$ , sinx neither achieves 1 nor achieves -1.

Hence

$a=1$ or

$a=-1$ for the equation

$sin\theta=a$ to have only one solution in the given interval of

$\theta$ .

$\displaystyle x,y\epsilon \left [ 0,2\pi \right ]$ and

$\displaystyle \sin x+\sin y=2$ then the value of

$x+y$ is

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$\displaystyle \pi$
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$\displaystyle \frac{\pi}{2}$
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$\displaystyle3 \pi$
0%
None of these

Explanation

$sinx + siny = 2$

but we know that maximum value of

$sinx$ is 1

hence only solution is

$sinx = 1$ and

$siny = 1$

so in

$0 \leq x \leq 2\pi, 0 \leq y \leq 2\pi$

solution is

$x = \dfrac{\pi}{2}, y = \dfrac{\pi}{2}$

therefore,

$x + y = \pi$

The number of solutions of the equation

$\displaystyle \cos 6x+\tan ^{2}x+\cos 6x\cdot \tan ^{2}x= 1$ in the interval

$\displaystyle \left [ 0 ,2\pi \right ]$ is

0%
4
0%
5
0%
6
0%
7

Explanation

$cos6x + tan^2x +cos6x.tan^2x = 1$

$\Rightarrow cos6x(1 + tan^2x) = (1 - tan^2x) \Rightarrow cos6x = \dfrac{1 - tan^2x}{1 + tan^2x}$

$\Rightarrow cos6x = cos2x$

$[\because \sin 2A=\dfrac {2 \tan A}{1+ \tan^2A}$ ,

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

Threrefore general solution is,

$6x = 2n\pi \pm 2x$ , where

$n$ is any integer

$x = \dfrac{n\pi}{2}$ and

$x = \dfrac{n\pi}{4}$

Hence solutions in the given interval are,

$x = 0, \dfrac{\pi}{4}, \dfrac{\pi}{2}, \dfrac{3\pi}{4}, \pi, \dfrac{3\pi}{4}, 2\pi$ , total

$7$ solutions.

The number of solutions of the equation

$\tan x+\sec x=2\cos x$ lying in the interval

$\left [ 0, 2\pi \right ]$ is

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

Explanation

$tanx + secx = 2 cosx$

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

$2 sin^2x + sinx -1 = 0$

$(2 sinx - 1) (sinx + 1) = 0$

but

$sinx = -1 \Rightarrow x = \dfrac{3\pi}{2}$ , which does not satisfy the given equation.

$sinx = \dfrac{1}{2} = sin\dfrac{\pi}{6}$

therefore general solution is,

$x = n\pi + (-1)^n.\dfrac{\pi}{6}$

$x = ......, \dfrac{\pi}{6}, \dfrac{5\pi}{6}, ......$

hence number of solution in the given interval is 2.

The number of real solutions of the equation

$\displaystyle \cos ^{7}x+\sin ^{5}x=1, 0\leq x\leq 2\pi$ , is

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

The total number of solutions of

$\displaystyle \cos x=\sqrt{1-\sin 2x}$ in

$\displaystyle \left [ 0,2\pi \right ]$ is equal to

0%
2
0%
3
0%
5
0%
None of these

Explanation

$cosx = \sqrt{1 - sin2x}$

$\Rightarrow cosx = \sqrt{ cos^2x + sin^2x - 2 sinx. cosx}$

$\Rightarrow cosx = \pm (cosx - sinx)$ , provided

$cosx \geq 0$

$\Rightarrow sinx = 0$ and

$tanx = 2$

Thus solutions in

$x \epsilon [0, 2\pi]$ with taking care that

$cosx \geq 0$ are,

$x = 0, tan^{-1}2, 2\pi$

$\sin \:2 \theta= \cos \:3\theta .$ The number of elements for the set

$\theta$ in

$0\leq \theta \leq 2\pi .$

0%
2
0%
5
0%
6
0%
12

Explanation

Given,

$sin2\theta = cos3\theta$

$\Rightarrow cos3\theta = cos(\dfrac{\pi}{2} - 2\theta)$

Therefore general solution is,

$3\theta = 2n\pi \pm (\dfrac{\pi}{2} - 2\theta)$ , where n is any integer

$\Rightarrow \theta = 2n\pi - \dfrac{\pi}{2}, (4n + 1)\dfrac{\pi}{10}$

thus solution in given interval is,

$\theta = \dfrac{\pi}{10}, \dfrac{\pi}{2}, \dfrac{9\pi}{10}, \dfrac{13\pi}{10}, \dfrac{3\pi}{2}, \dfrac{17\pi}{10}$

Hence, option 'C' is correct.

Find the value of

$x$ for which

$f\left ( x \right )=\sqrt{\sin x-\cos x}$ is defined,

$x\epsilon \left [ 0, 2\pi \right ]$

0%
$\displaystyle x\epsilon \left [ \frac{\pi }{8}, \frac{5\pi }{8} \right ]$
0%
$\displaystyle x\epsilon \left [ \frac{\pi }{4}, \frac{3\pi }{4} \right ]$
0%
$\displaystyle x\epsilon \left [ \frac{\pi }{6}, \frac{5\pi }{6} \right ]$
0%
$\displaystyle x\epsilon \left [ \frac{\pi }{4}, \frac{5\pi }{4} \right ]$

Explanation

$f\left ( x \right )=\sqrt{\sin x-\cos x}$ is defined if

$\sin x\geq \cos x$
From the graph,

$\sin x\geq \cos x$ , for

$\displaystyle x\epsilon \left [ \frac{\pi }{4}, \frac{5\pi }{4} \right ]$

0%
if both the statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1
0%
if both the statements are TRUE and STATEMENT 2 is NOT the correct explanation of STATEMENT 1
0%
if STATEMENT 1 is TRUE and STATEMENT 2 is FALSE
0%
if STATEMENT 1 is FALSE and STATEMENT 2 is TRUE

Explanation

$n|\sin (x)|=m|\cos (x)|$
$\dfrac{n}{m}=|cot(x)|$
Or
$\dfrac{m}{n}=|\tan (x)|$
Or
$\tan (x)=\dfrac{m}{n}$
Hence
$x=\tan ^{-1}(\dfrac{m}{n})$ and $x=\pi+\tan ^{-1}(\dfrac{m}{n})$ .
And
$\tan (x)=\dfrac{-m}{n}$
Hence
$x=-\tan ^{-1}(\dfrac{m}{n})$ and $x=\pi-\tan ^{-1}(\dfrac{m}{n})$ .
Now if $n,m\neq 0$ we therefore get 4 solutions in the interval of $[0,2\pi]$ independent of the values of m and n.

CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 3 - 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 3 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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