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