If $$\alpha \le 2\sin^{-1}x+\cos^{-1}x \le \beta$$, then

$$\alpha =\dfrac{-\pi}{2}, \beta =\dfrac{\pi}{2}$$

$$\alpha =\dfrac{-\pi}{2}, \beta =\dfrac{3\pi}{2}$$

If $$ \sin ^{-1} x = y $$ , then

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

MCQ Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 7

$\tan^{-1}x=\dfrac{\pi}{10}$ for some

$x\in R$ , then the value of

$\cot^{-1}x$ is

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

Explanation

(B) is the correct answer.

We know that

$\tan^{-1}x+\cot^{-1}x=\dfrac{\pi}{2}$ .

$\Rightarrow \cot^{-1}x=\dfrac {\pi}{2}-\tan^{-1}x$

$\Rightarrow \cot^{-1}x=\dfrac{\pi}{2}-\dfrac{\pi}{10}=\dfrac{4\pi}{10}=\dfrac{2\pi}{5}$

The value of

$\sin^{-1}\left\{\cos \left( \dfrac{43\pi}{5}\right)\right\}$ is

0%
$\dfrac{3\pi}{5}$
0%
$\dfrac{-7\pi}{5}$
0%
$\dfrac{\pi}{10}$
0%
$-\dfrac{\pi}{10}$

Explanation

(D) is the correct answer.

We have

$\sin^{-1}\left( \cos \dfrac{40\pi +3\pi}{5}\right)$

$=\sin^{-1}\left\{\cos \left( 8\pi+\dfrac{3\pi}{5}\right\}\right\}$

$=\sin^{-1}\left( \cos \dfrac{3\pi}{5}\right)$

$=\sin^{-1}\left\{ \sin \left( \dfrac {\pi}{2}-\dfrac{3\pi}{5}\right)\right\}$

$=\sin^{-1}\left( \sin \left( -\dfrac{\pi}{10}\right)\right)$

$=-\dfrac{\pi}{10}$

The value of

$\cot (\sin^{-1}x)$ is

0%
$\dfrac{\sqrt{1+x^2}}{x}$
0%
$\dfrac{x}{\sqrt{1+x^2}}$
0%
$\dfrac{1}{x}$
0%
$\dfrac{\sqrt{1-x^2}}{x}$

Explanation

(D) is the correct answer.

Let

$\sin^{-1}x=\theta$ ,

Then

$\sin \theta =x$

$\Rightarrow \csc \theta =\dfrac 1x$

$\Rightarrow \csc^2 \theta =\dfrac {1}{x^2}$

$\Rightarrow 1+\cot^2\theta =\dfrac {1}{x^2}$

$\Rightarrow \cot \theta =\dfrac {\sqrt{1-x^2}}{x}$

The domain of

$y=\cos^{-1}(x^2-4)$ is

0%
$[3, 5]$
0%
$[0, \pi]$
0%
$[-\sqrt 5, -\sqrt 3] \cap [ -\sqrt 5, \sqrt 3]$
0%
$[ -\sqrt 5, -\sqrt 3] \cup [ \sqrt 3, 5]$

Explanation

(D) is the correct answer.

Given

$y=\cos^{-1}(x^2-4) \Rightarrow \cos y =x^2-4$

$\therefore \ -1 \le x^2 -4 \le 1$ ( since

$-1 \le \cos y \le 1)$

$\Rightarrow 3 \le x^2 \le 5$

$\Rightarrow \sqrt 3 \le |x| \le \sqrt 5$

$\Rightarrow x \in [ -\sqrt 5, -\sqrt 3] \cup [ \sqrt 3, \sqrt 5]$

Hence the domain of

$y$ is

$[ -\sqrt 5, -\sqrt 3] \cup [ \sqrt 3, 5]$

The domain of the function

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

0%
$[0, 1]$
0%
$(0, 1)$
0%
$[-1, 1]$
0%
$\phi$

Explanation

Given

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

$\therefore\ -1 \le -x^2 \le 1$ ( since

$-1 \le \sin y \le 1)$

$\Rightarrow 1 \ge x^2 \ge -1$

$\Rightarrow 0 \le x^2 \le 1$

$\Rightarrow |x| \le 1$

$\Rightarrow-1 \le x \le 1$

Hence the domain is

$[-1, 1]$

The value of

$\sin (2\sin^{-1}(0.6))$ is

0%
$0.48$
0%
$0.96$
0%
$1.2$
0%
$\sin 1.2$

Explanation

(B) is the correct answer.

Let

$\sin^{-1}(0.6) =\theta \Rightarrow\sin \theta =0.6$ .

$\therefore\ \sin (2\sin^{-1}(0.6))\\=\sin ( 2\theta )\\=2\sin \theta \cos \theta \\=2 \sin\theta\sqrt{1-\sin^2\theta}\\=2(0.6)(\sqrt{1-{0.6}^2})\\=2(0.6)(0.8)=0.96$

The value of the expression

$\sin [ \cot^{-1}\{ \cos (\tan^{-1}1)\}]$ is

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

Explanation

(D) is the correct answer.

Let

$P=\sin [ \cot^{-1}\{ \cos (\tan^{-1}1)\}]$

$\\=\sin \left[ \cot^{-1}\left( \cos \dfrac {\pi}{4}\right) \right] \qquad\left[\because \tan^{-1}1=\dfrac{\pi}4\right]\\=\sin \left[ \cot^{-1}\dfrac{1}{\sqrt 2}\right] \qquad\left[\because \cos\dfrac{\pi}4=\dfrac1{\sqrt2}\right]\\$

Let,

$\cot^{-1}\dfrac1{\sqrt2}=y\Rightarrow \cot y=\dfrac1{\sqrt2}$

$\\ \Rightarrow \sin y=\sqrt{\dfrac23}\Rightarrow y=\sin^{-1}\sqrt{\dfrac23}$

$\therefore P=\sin \left[ \sin^{-1}\sqrt{\dfrac 23}\right] =\sqrt{\dfrac 23}$

The value of

$\tan^2 ( \sec^{-1}2)+\cot^2 (\csc^{-1}3)$ is

0%
$5$
0%
$11$
0%
$13$
0%
$15$

Explanation

(B) is the correct answer.

We have

$\tan^2 ( \sec^{-1}2) \cot^2 ( \csc^{-1}3)$

$=\sec^2 (\sec^{-1}2)-1+\csc^2 (\csc^{-1}3)-1\qquad[\because \tan^2\theta=\sec^2\theta-1,\ \ \ \cot^2\theta=\csc^2\theta-1]$

$=(2)^2-1+(3)^2-1=11$

$\alpha \le 2\sin^{-1}x+\cos^{-1}x \le \beta$ , then

0%
$\alpha =\dfrac{-\pi}{2}, \beta =\dfrac{\pi}{2}$
0%
$\alpha =0, \beta =\pi$
0%
$\alpha =\dfrac{-\pi}{2}, \beta =\dfrac{3\pi}{2}$
0%
$\alpha =0, \beta =2\pi$

Explanation

(B) is the correct answer.

We know that

$\dfrac {-\pi}{2}\le \sin^{-1}x \le \dfrac {\pi}{2}\\$

$\Rightarrow \dfrac {-\pi}{2}+\dfrac {\pi}{2} \le \sin^{-1}x + \dfrac {\pi}{2} \le \dfrac {\pi}{2}+\dfrac {\pi}{2}\qquad [$ Addidng

$\dfrac{\pi}2 ]$

$\\\Rightarrow 0 \le \sin^{-1}x+ (\sin^{-1}x+\cos^{-1}x) \le \pi\qquad\left[\because(\sin^{-1}x+\cos^{-1}x)= \dfrac{\pi}2 \right]\\$

$\Rightarrow 0 \le 2\sin^{-1}x+\cos^{-1}x \le \pi$

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

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

Explanation

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

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

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

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

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

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

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

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

$2x^{2} -x = 0 \\x(2x -1) = 0\\ \Rightarrow x = 0 , x = \dfrac{1}{2}$

$\dfrac{1}{2}$ does not satisfy the equation

$\therefore x = 0$ is the only solution

Choose the correct answer

$\cos ^{-1} ( \cos \dfrac{7\pi }{6})$ is equal to

0%
$\dfrac{7\pi }{6}$
0%
$\dfrac{5\pi }{6}$
0%
$\dfrac{\pi }{3}$
0%
$\dfrac{\pi }{6}$

Explanation

$\cos^{-1} \cos\left ( \dfrac{7\pi }{6} \right )$

$\cos^{-1} \left ( \cos\left ( \pi + \dfrac{\pi }{6} \right ) \right ) = \cos^{-1} \left ( - \cos\dfrac{\pi}{6} \right )$

$=\pi- \cos^{-1} \left ( \cos\dfrac{\pi }{6} \right )$

$=\pi-\dfrac{\pi}{6}$

$= \dfrac{5\pi }{6}$

Choose the correct answer :

$\sin \left ( \dfrac{\pi }{3} - \sin^{-1}\left ( -\dfrac{1}{2} \right ) \right )$ is equal to

0%
$\pi$
0%
$- \dfrac{\pi }{2}$
0%
$1$
0%
$2 \sqrt{3}$

Explanation

$\sin \left [ \left ( \dfrac{\pi}{3} - \sin^{-1} \left ( \dfrac{-1}{2} \right ) \right ) \right ]$

$\left [ \sin \left [ \left ( \dfrac{\pi}{3} - \sin^{-1} \left ( \dfrac{-1}{2} \right ) \right ) \right ] \right ]$

$\sin \left ( \dfrac{\pi }{3} + \sin^{-1}\left ( \sin \dfrac{\pi }{6} \right ) \right )$

$\left ( as \sin^{-1} \left ( \dfrac{-1}{2} \right ) = \sin^{-1}\dfrac{1}{2} and \sin \dfrac{\pi }{6} = \dfrac{1}{2} \right )$

$= \sin \left ( \dfrac{\pi }{3} + \dfrac{\pi }{6} \right )$

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

$\sin ^{-1} x = y$ , then

0%
$0 \leq y \leq \pi$
0%
$- \dfrac{\pi}{2} \leq y \leq \dfrac{\pi }{2}$
0%
$0 < y < \pi$
0%
$- \dfrac{\pi}{2} < y < \dfrac{\pi }{2}$

Explanation

pv of

$\sin^{-1} x$ is

$\left [ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right ]$

$\therefore (B) - \dfrac{\pi}{2} \leq y \leq \dfrac{\pi }{2}$ is correct

$\tan^{-1} \sqrt{3} - \sec^{-1} (-2)$ is equal to

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

Explanation

$\tan^{-1} \sqrt{3} = \dfrac{\pi }{3} \epsilon \left [ \dfrac{-\pi }{2} ,\dfrac{\pi }{2} \right ]$

$\sec^{-1} (-2) = \pi - \dfrac{\pi }{3} = \dfrac{2 \pi}{3} \epsilon [0, \pi]$

$\therefore \tan^{-1} \sqrt{3} - sec^{-1} (-2) = \dfrac{\pi}{3} - \dfrac{2 \pi}{3}$

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

$\therefore$ Option (B) is correct .

Multiple choice Questions :

$2\sin \left ( \cos^{-1}\left ( \dfrac{-4}{5} \right ) \right )\times \cos \cos^{-1} \left ( \dfrac{-4}{5} \right )$

0%
$\dfrac{24}{25}$
0%
$- \dfrac{24}{25}$
0%
$- \dfrac{6}{25}$
0%
$\dfrac{- 6}{25}$

Explanation

$2\sin \left ( \cos^{-1}\left ( \dfrac{-4}{5} \right ) \right )\times \cos \cos^{-1} \left ( \dfrac{-4}{5} \right )$

$= 2 \sqrt{1-\left ( \dfrac{-4}{5} \right )^{2}} \times\left ( \dfrac{-4}{5} \right )$

$= -\dfrac{8}{5} \sqrt{\dfrac{9}{25}} = \dfrac{-8}{5} \times\dfrac{3}{5} = \dfrac{-24}{25}$

Multiple choice Questions :

$\sin^{-1}\left ( \dfrac{4}{5} \right ) + \cos^{-1}\left ( \dfrac{4}{5} \right )=$

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

Explanation

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

$\sin^{-1}\left ( \dfrac{4}{5} \right ) + \cos^{-1}\left ( \dfrac{4}{5} \right )= \dfrac{\pi}{2}$

Multiple choice Questions :

$\cos^{-1} \cos \left ( \dfrac{4 \pi}{3} \right ) =$

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

Explanation

$\cos^{-1} \cos \left ( \dfrac{4 \pi}{3} \right ) = \cos^{1}\cos \left ( \pi + \dfrac{\pi}{3} \right )$

$= \cos^{-1} \left [ - cos \dfrac{\pi}{3} \right ]$

$= \pi - cos^{-1} \left ( \cos\dfrac{\pi}{3} \right ) = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$

Multiple choice Questions :
The value of

$\tan^{-1} (\tan 5)$

0%
$2 \pi$
0%
$5 - 2 \pi$
0%
$5$
0%
$\dfrac{2 \pi}{3}$

Explanation

Let

$2\pi - 5 = \theta$

$\Rightarrow 5 = 2 \pi - \theta$

$\tan 5 = \tan ( 2 \pi - \theta )$

$\tan^{-1} (\tan 5) = \tan^{-1} \tan ( 2 \pi - \theta)$

$= \tan^{-1}[-\tan \theta]=-\theta$

$\tan ^{-1}\left ( \dfrac{x}{y} \right ) - \tan^{-1} \dfrac{x - y}{x + y}$ is equal to

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

Explanation

$\tan^{-1} \left ( \dfrac{x}{y} \right ) - \left ( \tan^{-1}\dfrac{\frac{x}{y}-1}{1+\dfrac{x}{y}} \right )$

$=\tan^{-1} \left ( \dfrac{x}{y} \right )-\tan ^{-1} \left ( \dfrac{x}{y} \right )+ \tan^{-1} (1)$

$\tan^{-1} 1 = \dfrac{\pi }{4}$

Multiple choice Questions :
If a > b > c ,

$\cot^{-1} \left ( \dfrac{1 + ab}{a - b} \right ) + \cot^{-1}\left ( \dfrac{1 + bc}{b - c} \right ) + \cot^{-1} \left ( \dfrac{1 + ac}{c - a} \right )$

0%
$0$
0%
$\pi$
0%
$2 \pi$
0%
$\dfrac{\pi}{2}$

Explanation

$\tan ^{-1}\left ( \dfrac{a - b}{1 + ab} \right ) + \tan^{-1} \left ( \dfrac{b - c}{1 + bc} \right ) + \tan^{-1} \left ( \dfrac{c - a}{1 + ac} \right )$

$= \tan^{-1} a - \tan^{-1} b + \tan^{-1} b - \tan^{-1} c + \pi - \tan^{-1} a + \tan^{-1} c$

$= \pi \ \ \ \ \ \ \ \ as \tan^{-1} \left ( \dfrac{c - a}{1 + ac} \right ) = \pi - \tan^{-1} \left ( \dfrac{a - c}{1 + ac} \right )$

$\sin^{-1}\left(\dfrac{1}{2}\right)=x$ , then general value

$x$ $ is:

0%
$2n\pi \pm \dfrac{\pi}{6}$
0%
$\dfrac{\pi}{6}$
0%
$n\pi \pm \dfrac{\pi}{6}$
0%
$n\pi +(-1)^n\dfrac{\pi}{6}$

Explanation

Given,

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

$\Rightarrow \sin x=\dfrac{1}{2}=\sin \dfrac{\pi}{6}$

$\Rightarrow x=\dfrac{\pi}{6}$

$\therefore$ General value of

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

Hence, option

$(d)$ is correct.

$\tan^{-1}(1)+\cos^{-1}(\dfrac{1}{\sqrt{2}})=\sin^{-1}x$ , then value of

$x$ is

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

Explanation

$\tan^{-1}(1)+\cos^{-1}\left(\dfrac{1}{\sqrt{2}}\right)=\sin^{-1}x$

$\Rightarrow \sin^{-1}x=\dfrac{\pi}{4}+\dfrac{\pi}{4}$

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

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

$\Rightarrow x=1$

Hence, option

$(c)$ is correct.

$2\tan(\tan^{-1}x+\tan^{-1}x^3)$ is:

0%
$\dfrac{2x}{1-x^2}$
0%
$1+x^2$
0%
$2x$
0%
none of these

Explanation

$2\tan (\tan^{-1}x+\tan^{-1}x^{3})$

$=2\tan \left\{\tan^{-1}\left(\dfrac{x+x^{3}}{1-x\times x^{3}}\right)\right\}$

$=2\tan \left\{\tan^{-1}\left(\dfrac{x(1+x^{2})}{1-x^{4}}\right)\right\}$

$=2\tan \left\{\tan^{-1}\dfrac{x(1+x^{2})}{(1-x^{2})(1+x^{2})}\right\}$

$=2\tan \dfrac{x}{1-x^{2}}=\dfrac{2x}{1-x^{2}}$

Hence, option

$(a)$ is correct.

Value of

$\sin^{-1}(\dfrac{\sqrt{3}}{2})+2\cos^{-1}(\dfrac{\sqrt{3}}{2})$ is:

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

Explanation

$\sin^{-1}\left(\dfrac{\sqrt{3}}{2}\right)+2\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)$

$=\left[\sin^{-1}\left(\dfrac{\sqrt{3}}{2}\right)+\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)\right]+\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)$

$=\dfrac{\pi}{2}+\dfrac{\pi}{6}=\dfrac{2\pi}{3}\left[\because \sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}\right]$

Hence, option

$(c)$ is correct.

$\cot^{-1}x+\tan^{-1}\dfrac{1}{3}=\dfrac{\pi}{2}$ then

$x$ is:

0%
$1$
0%
$3$
0%
$\dfrac{1}{3}$
0%
none of these

Explanation

$\cot^{-1}(x)+\tan^{-1}\left(\dfrac{1}{3}\right)=\dfrac{\pi}{2}$

$\Rightarrow \cot^{-1}x=\dfrac{\pi}{2}-\tan^{-1}\left(\dfrac{1}{3}\right)$

$\Rightarrow \cot^{-1}x=\cot^{-1}\dfrac{1}{3}$

On comparing

$x=\dfrac{1}{3}$

Hence, option

$(c)$ is correct.

$\tan^{-1}(3x)+\tan^{-1} 2x=\dfrac{\pi}{4}$ , then

$x$ is:

0%
$\dfrac{1}{6}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{10}$
0%
$\dfrac{1}{2}$

Explanation

$\tan^{-1}(3x)+\tan^{-1}(2x)=\dfrac{\pi}{4}$

$\Rightarrow \tan^{-1}\left\{\dfrac{3x+2x}{1-3x\times 2x}\right\}=\dfrac{\pi}{4}$

$\Rightarrow \left(\dfrac{5x}{1-6x^{2}}\right)=\tan \dfrac{\pi}{4}$

$\Rightarrow \dfrac{5x}{1-6x^{2}}=1$

$\Rightarrow 1-6x^{2}=5x$

$\Rightarrow 6x^{2}+5x-1=0$

$\Rightarrow 6x^{2}+6x-x-1=0$

$\Rightarrow 6x(x+1)-1(x+1)=0$

$\Rightarrow (x+1)(6x-1)=0$

$\Rightarrow x=-1, x=\dfrac{1}{6}$

Hence, option

$(a)$ is correct.

lf the equation

$\displaystyle \sin^{-1}(x^{2}+x+1)+\cos^{-1}(\lambda x+1)=\frac{\pi}{2}$ has exactly two solutions, then

$\lambda$ can not have the integral value(s)

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

Explanation

$\displaystyle \sin ^{-1}(x^{2}+x+1)+\cos ^{-1}(\lambda x+1)=\frac{\pi }{2}$

$\displaystyle \cos ^{-1}(\lambda x+1)=\frac{\pi }{2} -\sin ^{-1}(x^{2}+x+1)$
Taking

$\cos$ on both sides,

$\lambda x+1=x^{2}+x+1$

$x^{2}+(1-\lambda )x=0$

$x(x+1-\lambda )=0$

$x=0$ or

$\lambda =x+1$

$\sin ^{-1}(x^{2}+x+1)$ is defined for

$-1 \leq x^{2}+x+1 \leq 1$

$x^{2}+x+2 \geq 0$ is always true.

$x(x+1) \leq 0 \Rightarrow x\in [-1,0]$

$\Rightarrow x+1 \in [0,1]$

$\lambda \in [0,1]$

$\lambda \neq -1,2$

Also when

$\lambda =1$ , there is only one solution to the given equation i.e.,

$x=0$
So,

$\lambda\neq 1$ .

Assertion(A):

$\cos^{-1}x$ and

$\tan^{-1}x$ are positive for all positive real values of

$x$ in their domain.
Reason(R): The domain of

$f(x)=\cos^{-1}x+\tan^{-1}x$ is

$[-1, 1].$

0%
Both A and R are true and R is the correct explanation of A
0%
Both A and R are true but R is not correct explanation of A
0%
A is true but R is false
0%
A is false but R is true

Explanation

Assertion :

$\cos ^ {-1} x$ range is

$\left[0,\pi \right]$ and domain is

$[-1,1]$

$\therefore \,\cos^{-1}x>0\,$ for all

$x>0$

$\tan ^ {-1} x$ range is

$\left(\dfrac{ -\pi }2 ,\dfrac \pi 2\right)$ and range is

$R$

$\forall x >0 \implies \tan ^ {-1} x >0$

Reason :

The domain of

$\cos ^ {-1} x$ is

$[-1,1]$

The domain of

$\tan ^ {-1} x$ is

$R$

So The domain of

$\cos ^ {-1}x +\tan ^ {-1}x$ is

$R \cap [-1,1]$

$\implies [-1,1]$

$\sin^{-1}|\sin x|=\sqrt{\sin^{-1}|\sin x|}$ then

$x=$

0%
$n\pi-1$
0%
$n\pi$
0%
$n\pi+1$
0%
$n\displaystyle \frac{\pi}{2}+1$

Explanation

Case - 1

$|sin x|=sin x \ \ x\epsilon [2n\pi , (2n+1)\pi ]$

$sin ^{-1} sin x =\sqrt{sin ^{-1}sin x}$

$x=\sqrt{x}$

$\sqrt{x}=1\ \ or \ \ x=0$

$x=1$
Case - 2

$|sin x|=-sin x\ \ x\epsilon [(2n-1)\pi , 2n\pi ]$

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

$-x=\sqrt{-x}$

$x=-1 , x=0$
So,

$x=n\pi -1 \ for\ n=0$

$x=n\pi\ for\ n=0$

$x=n\pi -1 \ for\ n=0$

The number of solutions of:

$\displaystyle \sin^{-1}(1+b+b^{2}+\ldots.\infty)+\cos^{-1}(a-\frac{a^{2}}{3}+\frac{a^{3}}{9}+\ldots\infty)=\frac{\pi}{2}$

0%
$1$
0%
$2$
0%
$3$
0%
$\infty$

Explanation

$\displaystyle \sin ^{ -1 }{ \theta }$ are defined for

$\displaystyle \theta \in \left[ -1,1 \right]$

$\therefore$ For

$\displaystyle \sin ^{ -1 }{ \left( 1+b+{ b }^{ 3 }+...\infty \right) }$

$\left| b \right| <1$

Hence

$\displaystyle 1,b,{ b }^{ 2 },...\infty$ is

$G.P.$ of common ratio

$b$

$\displaystyle \therefore 1+b+{ b }^{ 2 }+...+\infty =\frac { 1 }{ 1-b }$

Similarly for

$\displaystyle \cos ^{ -1 }{ \left( a-\frac { { a }^{ 2 } }{ 3 } +\frac { { a }^{ 3 } }{ 9 } +...\infty \right) }$

$\displaystyle\left| \frac { a }{ 3 } \right| <1$

Hence

$\displaystyle 1,-\frac { { a }^{ 2 } }{ 3 } ,\frac { { a }^{ 3 } }{ 9 } ,...\infty$ is

$G.P.$ of common ration

$\displaystyle \left( \frac { -a }{ 3 } \right)$

$\displaystyle \therefore a-\frac { { a }^{ 2 } }{ 3 } +\frac { { a }^{ 3 } }{ 9 } +...\infty =\frac { a }{ 1+\frac { a }{ 3 } } =\frac { 3a }{ 3+a }$

Now, as

$\displaystyle \sin ^{ -1 }{ \left( x \right) } +\cos ^{ -1 }{ \left( x \right) } =\frac { \pi }{ 2 }$

$\displaystyle \therefore \frac { 1 }{ 1-b } =\frac { 3a }{ 3+a }$

$\displaystyle \Rightarrow 3+a=3a-3ab$

$\displaystyle \Rightarrow 2a-3=3ab$

This has infinite solution.

$(\tan^{-1}x)^{2}+(\cot^{-1}x)^{2} = \displaystyle \frac{5\pi^{2}}{8}$ , then

$x=$

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

Explanation

Let

$tan^{-1}x=y$
Therefore

$y^2+(\frac{\pi}{2}-y)^{2}=\frac{5\pi^2}{8}$

$2y^2-\frac{2\pi y}{2}+\frac{\pi^2}{4}=\frac{5\pi^2}{8}$

$2y^2-\pi y-\frac{3\pi^2}{8}=0$

$y=\frac{-\pi}{4}$ and

$y=\frac{3\pi}{4}$
Hence

$tan^{-1}(x)=\frac{-\pi}{4}$

$x=-1$ and

$tan^{-1}(x)=\frac{3\pi}{4}$

$x=-1$
However only

$x=-1$ satisfies the above quadratic equation.

$cos^{-1} \left (\sqrt{\dfrac{a-x}{a-b}} \right)$ =

$sin^{-1} \left (\sqrt{\dfrac{x-b}{a-b}}\right)$ is possible if

0%
$a>x>b$ or $a< x < b$
0%
$a=x=b$
0%
$a > b$ and x takes any value
0%
$a < b$ and x takes any value

Explanation

$\cos ^{-1}x , x \in [-1,1]$

$\sin ^{-1}x , x\in [-1,1]$

$\displaystyle -1 \leq \sqrt{\frac{a-x}{a-b}} \leq 1$

$\displaystyle 0 \leq \sqrt{\frac{a-x}{a-b}} \leq 1$

$0 \leq a-x \leq a-b$

$0\geq x-a \geq b-a$ [on multiplying by (-ve) sing direction of equality changes]

$a \geq x\geq b$
and

$\displaystyle 0\leq \sqrt{\frac{x-b}{a-b}}\leq 1$

$0\leq x-b \leq a-b$

$b \leq x\leq a$

$x\epsilon [b,a]$

The solution set of the equation

$\tan^{-1}x -\cot^{-1}x =\cos^{-1}(2-x)$ is

0%
$(0,1)$
0%
$(-1,1)$
0%
$[1,3)$
0%
$(1,3)$

Explanation

Given,

$\tan ^{-1}x-\cot ^{-1}x=\cos ^{-1}(2-x) x\in [1,3]$

$2\tan ^{-1}x-\dfrac{\pi }{2}=\cos ^{-1}(2-x)$

$\left[\because \tan ^{-1}x+\cot ^{-1}x=\dfrac{\pi }{2}\right]$

$2 \tan ^{-1}x=\dfrac{\pi }{2}+\cos ^{-1}(2-x)$

Take

$\sin$ on both sides,

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

$\Rightarrow x^3-2x^{2}+3x-2=0$

$\Rightarrow x = 1$

So, set containing solution is

$[1,3)$ .

$\sin^{-1}\alpha+\sin^{-1}\beta+\sin^{-1}\gamma =\displaystyle \frac{3\pi}{2}$ , then

$\alpha\beta+\alpha\gamma+\beta\gamma$ is equal to :

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

Explanation

The above expression is true for

$\alpha=1$ ,

$\beta=1$ and

$\gamma=1$
Since

$\dfrac{-\pi}{2}\leq sin^{-1}(x) \leq \dfrac{\pi}{2}$
Hence

$\alpha\beta+\beta\gamma+\gamma\alpha$

$=(1)+(1)+(1)$

$=3$

The number of positive integral solutions of the equation

$tan^{-1}x+cot^{-1}y =tan^{-1} 3$ is :

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

Explanation

$cot^{-1}y=tan ^{-1}3-tan ^{-1}x$

$\displaystyle cot^{-1}y=tan ^{-1}(\dfrac{3-x}{1+3x})$
Take tan on both sides

$\displaystyle \dfrac{1}{y}=\dfrac{3-x}{1+3x}$

$\displaystyle y=\dfrac{1+3x}{3-x} \Rightarrow y > 0$
Put

$x=0 , y=\dfrac{1}{3}$

$\dfrac{1+3x}{3-x} > 0$

$x=1 , y=2$

$\dfrac{3x+1}{x=3} < 0$

$x=2 , y=7$

$x\epsilon (-\dfrac{1}{3} , 3)$
The integral values of

$x=0,1,2$

The domain of

$\displaystyle \mathrm{f}(\mathrm{x})=\cot^{-1}\left(\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}-[\mathrm{x}^{2}]}}\right)$ is
(

$[\;.]$ denotes the greatest integer function)

0%
$(0,\infty)$
0%
$\mathrm{R}-\{0\}$
0%
$\mathrm{R}-\{\mathrm x:\mathrm{x}\in \mathrm{Z}\}$
0%
$(-\infty,0)$

Explanation

$f(x)=\cot^{-1}\left ( \displaystyle \frac{x}{\sqrt{x^2-[x^2]}} \right )$
For

$f(x)$ to be defined,

$x^2-[x^2]>0$

$\therefore x^2$ should not be an integer
Hence,

$x$ should not be an integer.
Hence, Option C is the answer

$(tan^{-1} x)^2 +(cot ^{-1}x)^2=\displaystyle \frac{5 \pi^2}{8}$ , then

$x$ =

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

Explanation

We have,

$tan^{-1}x + cot^{-1}x =\displaystyle\frac{\pi}{2}$

The given equation can be written as :

$\displaystyle (tan^{-1}x + cot^{-1}x)^2 - 2tan^{-1}x\left ( \frac{\pi}{2}-tan^{-1}x \right )=\frac{5 \pi^2}{8}$

$\displaystyle 2(tan^{-1}x)^2 - 2\left ( \frac{\pi}{2} \right )tan^{-1}(x)- \frac{3 \pi ^2}{8}=0$

$\displaystyle tan^{-1}x=-\frac{\pi}{4} or \frac{3\pi}{4}$

Hence,

$tan^{-1}x =1$

$x=-1$

The number of integral solutions of

$sin^{-1}\sqrt{4x-x^{2}-3}+tan^{-1}\sqrt{x^{2}-3x+2}=\frac{\pi }{2}$ is

0%
zero
0%
infinite
0%
four
0%
None of these

Explanation

Given equation is

$\sin^{-1}\sqrt{4x-x^{2}-3}+\tan^{-1}\sqrt{x^{2}-3x+2}=\frac{\pi }{2}$

$\tan^{ -1 }\sqrt { x^{ 2 }-3x+2 }=\displaystyle\frac { \pi }{ 2 } -\sin^{ -1 }\sqrt { 4x-x^{ 2 }-3 }$

$\Rightarrow \tan^{ -1 }\sqrt { x^{ 2 }-3x+2 } =\cos^{ -1 }\sqrt { 4x-x^{ 2 }-3 }$

Since,

$\sqrt { x^{ 2 }-3x+2 } \ge 0$

$\Rightarrow \tan ^{ -1 }{ \sqrt { x^{ 2 }-3x+2 } } <\dfrac { \pi }{ 2 }$

Also,

$\sqrt { 4x-x^{ 2 }-3 } \ge 0$

$\Rightarrow 0<\cos^{ -1 }\sqrt { 4x-x^{ 2 }-3 } \le \dfrac { \pi }{ 2 }$

The value of

$sin^{-1}(sin2010^{0})+cos^{-1}(cos2010^{0})+tan^{-1}(tan2010^{0})$ is

0%
$\frac{\pi }{6}$
0%
$\frac{\pi }{3}$
0%
$\frac{2\pi }{3}$
0%
$\frac{5\pi }{6}$

Explanation

$2010^{0}$

$=11\pi+\frac{\pi}{6}$ ...(i)
Hence

$sin^{-1}(sin(2010^{0}))=sin^{-1}(sin(11\pi+\frac{\pi}{6}))$ ...(from(i))

$cos^{-1}(cos(2010^{0}))=cos^{-1}(cos(11\pi+\frac{\pi}{6}))$

$tan^{-1}(tan(2010^{0}))=tan^{-1}(tan(11\pi+\frac{\pi}{6}))$
Therefore

$sin^{-1}(sin(11\pi+\frac{\pi}{6}))+cos^{-1}(cos(11\pi+\frac{\pi}{6}))+tan^{-1}(tan(11\pi+\frac{\pi}{6}))$

$=-\frac{\pi}{6}+\pi-\frac{\pi}{6}+\frac{\pi}{6}$

$=\frac{5\pi}{6}$

The number of solutions of the equation

$1+x^{2}+2x\, sin\: (cos^{-1}y)=0$ is

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

Explanation

$1+x^{2}+2x sin cos ^{1-}y=0$

$\Rightarrow 1+x^{2}+2x\sqrt{1-y^{2}}=0$

$x^{2}+2x\sqrt{1-y^{2}}+1=0$

$x^{2}+2x\sqrt{1-y^{2}}+1-y^{2}=-y^{2}$

$(x+\sqrt{1-y^{2}})=-y^{2} \Rightarrow -y^{2} \geq 0$

$x+\sqrt{1-y^{2}}=0 \space and \space y=0 (x+\sqrt{1-y^{2}})^{2} \geq 0$

$\rightarrow y=0 , x=-1$ only one solution

The value of

$\displaystyle \sec^{-1}\left (\displaystyle \frac{1}{1-2x^{2}}\right)+4{\cos^{-1}}\sqrt{\displaystyle \frac{1+x}{2}}$ is equal to

0%
$\pi$
0%
$2\pi$
0%
$\dfrac{\pi}{2}$
0%
None of these

Explanation

Given

$\sec ^{-1} \left(\dfrac 1{1-2x^2}\right)+4\cos ^{-1}\sqrt {\dfrac {1+x}2}$

Let

$x=\cos \theta$

$\implies \sec ^{-1} \dfrac 1{1-2\cos ^2 x} +4\cos ^{-1}\left( \sqrt {\dfrac {1+\cos \theta }2}\right)$

$= \sec ^{-1}\left( \dfrac {-1}{\cos 2\theta }\right) +4\cos ^{-1} (\sqrt {\cos ^2\theta /2})$

$=\pi -\sec ^{-1} (\sec 2\theta )+4 \cos ^{-1} (\cos \theta/2)$

$=\pi -2\theta +4 \left(\dfrac \theta 2\right)$

$=\pi -2\theta +2\theta$

$=\pi$

The largest interval lying in

$\left ( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right )$ for which the function

$\left [ f(x)=4^{-x^{2}}+\cos^{-1}\left ( \dfrac{x}{2}-1 \right )+\log (\cos x) \right ]$ is defined, is-

0%
$[0,\pi ]$
0%
$\left ( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right )$
0%
$\left [- \dfrac{\pi }{4},\dfrac{\pi }{2} \right )$
0%
$\left [0,\dfrac{\pi }{2} \right )$

Explanation

For given function to be defined

$-1 \leq \cfrac{x}{2}-1 \leq 1$ and

$\cos x > 0$

$\Rightarrow 0 \leq \cfrac{x}{2} \leq 2$ and

$\cfrac{-\pi}{2} \leq x < \cfrac{\pi}{2}$

$\Rightarrow 0 \leq x \leq 4$ and

$\cfrac{-\pi}{2} \leq x < \cfrac{\pi}{2}$
Taking common of both we get

$x \in \left[0,\cfrac{\pi}{2}\right)$

The number of real solutions of

$tan^{-1} (\sqrt{x(x+1)}+sin^{-1} \displaystyle \sqrt{(x^{2}+x+1)}=\dfrac{\pi}{2}$ is

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

Explanation

From the above expression

$-1\leq\sqrt{x^2+x+1}\leq 1$ and

$x(x+1)\geq 0$

$\Rightarrow 0\le x^2+x+1\le1 \,and \,x^2+x+1\ge1$

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

$\Rightarrow x^2+x=0$

$\Rightarrow x(x+1)=0$
Hence there will be two solutions. One at

$x=-1$ and another at

$x=0$ .

$x> 0\,$ and

$\, cos^{-1}\left ( \dfrac{12}{x} \right )+cos^{-1}\left ( \dfrac{35}{x} \right )=\dfrac{\pi }{2},$ then x is

0%
$7$
0%
$39$
0%
$37$
0%
$-37$

Explanation

$cos^{-1}(a)+cos^{-1}(b)=\dfrac{\pi}{2}$

$\Rightarrow cos^{-1}(a)=\dfrac{\pi}{2}-cos^{-1}(b)$

$\Rightarrow cos^{-1}(a)=sin^{-1}(b)$
Therefore,

$a^2+b^2=1$
Substituting the value of a and b we get

$\dfrac{12}{x}^{2}+\dfrac{35}{x}^{2}=1$

$x^2=1369$

$x=37,-37$
However , since

$x>0$

$x=37$

The value of

$\sin^{-1}$

$\left \{ \tan\left ( \cos^{-1}\sqrt{\dfrac{2+\sqrt{3}}{4}}+\cos^{-1}\dfrac{\sqrt{12}}{4} -\text{cosec}^{-1}\sqrt{2}\right ) \right \}$ , is

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

Explanation

Given:

$E = \sin^{-1} \left(\tan \left(\cos^{-1} \dfrac {\sqrt {2 + \sqrt {3}}}{2} + \cos^{-1} \dfrac {\sqrt 3}{2} - \text{cosec}^{-1} \sqrt 2\right) \right)$

We know that,

$\cos 15^0 = \sqrt{\dfrac{2+\sqrt3}{4}}$

Therefore,

$\Rightarrow E= \sin^{-1} (\tan (15^0 + 30^0 - 45^0))$

$\Rightarrow E= \sin^{-1} (\tan0^0)$

$\Rightarrow E = \sin^{-1} 0 = 0$

$cosec ^{ -1 }\left(cosec (x) \right)$ and

$cosec\left(cosec ^{ -1 }(x) \right)$ are equal functions, then the maximum range of value of

$x$ is

0%
$\displaystyle\:\left [ -\frac{\pi }{2},-1\right ]\cup \left [ 1,\frac{\pi }{2} \right ]$
0%
$\displaystyle\:\left (-\frac{\pi }{2},-1\right )\cup \left (1,\frac{\pi }{2} \right )$
0%
$\displaystyle\:\left ( -\infty ,-1]\cup [1,\infty \right )$
0%
$\displaystyle\:\left ( -\infty ,-1)\cup (1,\infty \right )$

Explanation

The range of

$cosec^{-1}(cosec(x))$

$=\left[\dfrac{-\pi}{2},0\right)\cup\left(0,\dfrac{\pi}{2}\right]$
The range of

$cosec(cosec^{-1}(x))$

$=(-\infty,-1]\cup[1,\infty)$
Since it is given that both the functions are equal, then the range of value x common to both will be

$=\left[\dfrac{-\pi}{2},-1\right]\cup\left[1,\dfrac{\pi}{2}\right]$

$\left[ \sin ^{ -1 }{ \cos ^{ -1 }{ \sin ^{ -1 }{ \tan ^{ -1 }{ \theta } } } } \right] =1$ , where

$[.]$ denotes the greatest integer function, the

$\theta$ lies in the interval

0%
$[\tan { \sin { \cos { 1 } } } ,\sin { \tan { \cos { \sin { 1 } } } } ]$
0%
$[\sin { \tan { \cos { 1 } } } ,\tan { \sin { \cos { \sin { 1 } } } } ]$
0%
$[\tan { \sin { \cos { 1 } } } ,\tan { \sin { \cos { \sin { 1 } } } } ]$
0%
None of these

Explanation

We have

$\left[ \sin ^{ -1 }{ \cos ^{ -1 }{ \sin ^{ -1 }{ \tan ^{ -1 }{ \theta } } } } \right] =1$

$\Rightarrow 1\le \sin ^{ -1 }{ \cos ^{ -1 }{ \sin ^{ -1 }{ \tan ^{ -1 }{ \theta } } } }\le \displaystyle\frac { \pi }{ 2 }$

$\Rightarrow \sin { 1 } \le \cos ^{ -1 }{ \sin ^{ -1 }{ \tan ^{ -1 }{ \theta } } } \le 1$

$\Rightarrow \cos { \sin { 1 } } \ge \sin ^{ -1 }{ \tan ^{ -1 }{ \theta } } \ge \cos { 1 }$

$\Rightarrow \sin { \cos { \sin { 1 } } } \ge \tan ^{ -1 }{ \theta } \ge \sin { \cos { 1 } }$

$\Rightarrow \tan { \sin { \cos { \sin { 1 } } } } \ge \theta \ge \tan { \sin { \cos { 1 } } }$

$\displaystyle \:\cos ^{-1}x+\cos ^{-1}\left ( \frac{x}{2}+\frac{1}{2}\sqrt{3-3x^{2}} \right )$ is equal to

0%
$\displaystyle \:\frac{\pi }{3}$ for $x \epsilon \left [ \dfrac{1}{2},1 \right ]$
0%
$\displaystyle \:\frac{\pi }{3}$ for $x\epsilon \left [ 0,\dfrac{1}{2} \right ]$
0%
$\displaystyle \:2\cos ^{-1}x-\cos ^{-1}\dfrac{1}{2}$ for $x \epsilon \left [ \dfrac{1}{2},1 \right ]$
0%
$\displaystyle \:2\cos ^{-1}x-\cos ^{-1}\dfrac{1}{2}$ for $x \epsilon \left [ 0,\dfrac{1}{2}\right ]$

Explanation

$cos^{ -1 }x+cos^{ -1 }\left( x.\cfrac { 1 }{ 2 } +\cfrac { \sqrt { 3 } }{ 2 } \sqrt { 1-x^{ 2 } } \right) =cos^{ -1 }x+cos^{ -1 }\cfrac { 1 }{ 3 } -cos^{ -1 }x$
For

$\cfrac { 1 }{ 2 } \le x\le 1$

$cos^{ -1 }x+cos^{ -1 }\left( x.\cfrac { 1 }{ 2 } +\cfrac { \sqrt { 3 } }{ 2 } \sqrt { 1-x^{ 2 } } \right) =cos^{ -1 }x+\cfrac { \pi }{ 3 } -cos^{ -1 }x=\cfrac { \pi }{ 3 }$
And for

$0\le x\le \cfrac { 1 }{ 2 }$

$os^{ -1 }x+cos^{ -1 }\left( x.\cfrac { 1 }{ 2 } +\cfrac { \sqrt { 3 } }{ 2 } \sqrt { 1-x^{ 2 } } \right) =cos^{ -1 }x+cos^{ -1 }x-cos^{ -1 }\cfrac { 1 }{ 2 } =2cos^{ -1 }x-cos^{ -1 }\cfrac { 1 }{ 2 }$

The range of values of p for which the equation

$\sin \cos ^{-1}(\cos (\tan ^{-1}x))= p$ has a solution is

0%
$\left ( -\frac{1}{\sqrt{2}},\frac{2}{\sqrt{2}} \right )$
0%
$[0,1)$
0%
$\left ( \frac{1}{\sqrt{2}1} \right )$
0%
$(-1,1)$

Explanation

Here The range of

$cos(tan^{-1}(x))$ will be [-1,1] since the range of

$cos\theta$ is always

$[-1,1]$
Hence for

$cos(tan^{-1}(x))=1$

$sin(cos^{-1}(1))$

$=sin(0)$

$=0$
For

$cos(tan^{-1}(x))=0$

$sin(cos^{-1}(0))$

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

$=1$
For

$cos(tan^{-1}(x))=-1$

$sin(cos^{-1}(-1))$

$=sin(\pi)$

$=0$
Howere p=1 for

$x\rightarrow \infty$
Hence range of p=[0,1).

The solution set of the equation

$\displaystyle \sin^{-1} \sqrt{1-x} + \cos^{-1}x = \cot^{-1} \left ( \frac{\sqrt{1-x^{2}}}{x} \right )-\sin^{-1}x$

0%
$\displaystyle \left [ -1,\: 1 \right ] - \left \{ 0 \right \}$
0%
$\displaystyle \left ( 0, \: 1 \right ] \cup \left \{ -1 \right \}$
0%
$\displaystyle \left [ -1,\: 0 \right ) \cup \left \{ 1 \right \}$
0%
$\displaystyle \{ 1 \}$

Explanation

$\displaystyle \sin^{-1} \sqrt{1-x} + \cos^{-1}x = \cot^{-1} \left ( \frac{\sqrt{1-x^{2}}}{x} \right )-\sin^{-1}x$

$\Rightarrow \sin ^{ -1 } \sqrt { 1-x } +\dfrac { \pi }{ 2 } =\cot ^{ -1 } \left( \dfrac { \sqrt { 1-x^{ 2 } } }{ x } \right)$

$\Rightarrow \dfrac { \pi }{ 2 } -\cot ^{ -1 } \left( \dfrac { \sqrt { 1-x^{ 2 } } }{ x } \right) =-\sin ^{ -1 } \sqrt { 1-x }$

$\Rightarrow \tan ^{ -1 } \left( \dfrac { \sqrt { 1-x^{ 2 } } }{ x } \right) =-\sin ^{ -1 } \sqrt { 1-x }$

Thus is only true when,

$x = \{1\}$

CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 7 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 7 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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