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