If $ \arcsin x+ \arcsin y=\frac{\pi}{2}$, prove that $ x^2+y^2=1$. If $ \arcsin x+ \arcsin y=\frac{\pi}{2}$,
prove that $$ x^2+y^2=1$$
I tried taking sine of both the sides, I only come to this result: $$x^2 + y^2 -2x^2y^2 + 2xy\sqrt{(1-y^2)(1+x^2)}=1.$$

Using the identity $arcsiny + arccosy = \frac{\pi}{2}$
$$arcsinx + arcsiny=\frac{\pi}{2}$$
$$arcsinx + \frac{\pi}{2} -arcccosy=\frac{\pi}{2}$$
$$arcsinx =arccosy$$
Let $arcsinx = \theta =arccosy \iff cos\theta=y \land sin(\theta) =x$.
Using the pythagorean identity to show that $cos^2\theta +sin^2\theta=1 \iff y^2 + x^2 =1$
