Answer
$\dfrac{x-y}{\sqrt{1+y^2}\sqrt{1+x^2}}$
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 $\sin x=\dfrac{\tan x}{\sqrt {1+\tan^2 x}}$ and $\cos x=\dfrac{1}{\sqrt {1+\tan^2 x}}$
Now, $\sin (\tan^{-1}x -\tan^{-1}y)=\sin(p-q)$
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, $\sin (\tan^{-1}x -\tan^{-1}y)=\dfrac{x-y}{\sqrt{1+y^2}\sqrt{1+x^2}}$
