<s>[INST] 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: $$x^2 + y^2 -2x^2y^2 + 2xy\sqrt{(1-y^2)(1+x^2)}=1.$$
 [/INST] Let, $\sin^{-1}x=\alpha\iff x=\sin \alpha$ 
& $\sin^{-1}x=\beta\iff y=\sin\beta$ then we have$$\alpha+\beta=\frac{\pi}{2}$$
Now, we have 
$$x^2+y^2=(\sin\alpha)^2+(\sin\beta)^2$$
$$x^2+y^2=\sin^2\alpha+\sin^2\beta$$
Substituting $\beta=\frac{\pi}{2}-\alpha$, we get 
$$x^2+y^2=\sin^2\alpha+\sin^2\left(\frac{\pi}{2}-\alpha\right)$$
$$=\sin^2\alpha+\cos^2\alpha=1$$ 
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{x^2+y^2=1}}$$
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