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

The correct option is A $1-y^2$

We have, ${\sin }^{-1}x+{\sin }^{-1}y=\frac{\pi }{2}$

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

$\Rightarrow \sin \left({\sin }^{-1}x\right)=\sin (\frac{\pi }{2}-{\sin }^{-1}y)$, take sine both sides

$\Rightarrow x=\cos \left({\sin }^{-1}y\right)$, since $\sin ({90}^o-\theta )=\cos \theta$

$\Rightarrow x=\cos \left({\cos }^{-1}\sqrt{1-y^2}\right)=\sqrt{1-y^2}$

$\therefore x^2=1-y^2$, use basic rule of trigonometry