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 know that
${\sin }^{-1}t+{\cos }^{-1}t=\frac{\pi }{2}\forall t\in R\cdots \left(1\right)$
$\text{Given,}{\sin }^{-1}x+{\sin }^{-1}y=\frac{\pi }{2}$
$\Rightarrow {\sin }^{-1}x={\cos }^{-1}y\left[\text{From}\right(1\left)\right]$
$\Rightarrow {\sin }^{-1}x={\sin }^{-1}\sqrt{1-y^2}$
$\Rightarrow x=\sqrt{1-y^2}$
$\Rightarrow x^2=1-y^2$