Question

If $'x'$ takes negative permissible value, then $\sin^{-1} x$ is equal to

• A. $\cos^{-1} \sqrt {1-x^2}$
• B. $-\cos^{-1} \sqrt {1-x^2}$
• C. $\cos^{-1} \sqrt {x^2+1}$
• D. $\pi - \cos^{-1} \sqrt {1-x^2}$

Answer 1

Answer: (B)

Let $\sin ^{-1} x=y .$ Then, $x=\sin y$
Since, $-1 \leq x<0$, therefore $-\frac{\pi}{2} \leq \sin ^{-1} x<0$
$\Rightarrow -\frac{\pi}{2} \leq y<0$
Now, $\cos y=\sqrt{1-\sin ^{2} y}$
$\Rightarrow \cos y=\sqrt{1-x^{2}}$ for $0 \leq y \leq \pi$
But $\quad-\frac{\pi}{2} \leq y<0$
$\Rightarrow \frac{\pi}{2} \geq-y>0$
$\Rightarrow \cos (-y) =\sqrt{1-x^{2}} $
$\Rightarrow -y=\cos ^{-1} \sqrt{1-x^{2}} $
$ \Rightarrow y =-\cos ^{-1} \sqrt{1-x^{2}}$