Question

If $\sin^{-1} \, x + \sin^{-1} \,y= \frac{\pi}{2}$ , then $x^2$ is equal to

• A. $1 - y^2$
• B. $y^2$
• C. $0$
• D. $\sqrt{1 - y }$

Answer 1

Answer: (A)

We have, $\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1} x=\frac{\pi}{2}-\sin ^{-1} y$
$\Rightarrow \sin ^{-1} x=\cos ^{-1} y$
$\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]$
$\Rightarrow \sin ^{-1} x=\sin ^{-1}\left(\sqrt{1-y^{2}}\right)$
$\Rightarrow x=\sqrt{1-y^{2}} $
$\Rightarrow x^{2}=1-y^{2}$