Answer
$xy+\sqrt{1-x^2}\sqrt{1-y^2}$
Work Step by Step
Let us consider $p=\sin^{-1}x \implies x=\sin p$
and $q=\cos^{-1} y \implies y=\cos q$
We have $\sin x=\dfrac{\tan x}{\sqrt {1+\tan^2 x}}$ and $\cos x=\dfrac{1}{\sqrt {1+\tan^2 x}}$
Now, $\sin (\sin^{-1}x +\cos^{-1}y)=\sin(p+q)$
This gives: $\sin (p+q)=\sin p\cos q+\cos p \sin q$
or, $=(\dfrac{\tan p}{\sqrt {1+\tan^2 p}})(\dfrac{1}{\sqrt {1+\tan^2 q}})-(\dfrac{1}{\sqrt {1+\tan^2 p}}) (\dfrac{\tan q}{\sqrt {1+\tan^2 q}})$
or, $=(\dfrac{x}{\sqrt {1+x^2}})(\dfrac{1}{\sqrt {1+y^2}})-(\dfrac{1}{\sqrt {1+x^2}}) (\dfrac{y}{\sqrt {1+y^2}})$
Hence, $\sin (\sin^{-1}x +\cos^{-1}y)=xy+\sqrt{1-x^2}\sqrt{1-y^2}$
