Answer
$\tan (\sin^{-1}x+\cos^{-1}y)=\dfrac{xy+\sqrt{1-y^2}\sqrt{1-x^2}}{y\sqrt{1-x^2}-x\sqrt{1-y^2}}$
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 $\tan x=\dfrac{\sin x}{\sqrt {1-\sin^2 x}}$
Now, $\tan (\sin^{-1}x+\cos^{-1}y)=\tan (p+q)$
This gives: $\tan (p+q)=\dfrac{\tan p+\tan q}{1-\tan p\tan q}$
or, $=\dfrac{\dfrac{\sin p}{\sqrt {1-\sin^2 p}}+\dfrac{\sqrt{1-\cos^2 q}}{\cos q}}{1-(\dfrac{\sin p}{\sqrt {1-\sin^2 p}})(\dfrac{\sqrt{1-\cos^2 q}}{\cos q})}$
or, $=\dfrac{\dfrac{x}{\sqrt {1-x^2}}+\dfrac{\sqrt{1-y^2}}{y}}{1-(\dfrac{x}{\sqrt {1-x^2}})(\dfrac{\sqrt{1-y^2}}{y})}$
Hence, $\tan (\sin^{-1}x+\cos^{-1}y)=\dfrac{xy+\sqrt{1-y^2}\sqrt{1-x^2}}{y\sqrt{1-x^2}-x\sqrt{1-y^2}}$
