Answer
$\frac{x}{\sqrt {1 - x^{2} }}$
Work Step by Step
Given expression is-
$\tan(\sin^{-1}x)$
Assuming $\sin^{-1} x$ = $u$ , we get-
$\sin u$ = $x$, [$-\frac{\pi}{2}\leq u\leq \frac{\pi}{2}$]
From First Pythagorean identity-
$\cos u$ = + $\sqrt {1 - x^{2} }$
We know that,
$\tan u$ = $\frac{\sin u}{\cos u}$
i.e. $\tan u$ = $\frac{x}{\sqrt {1 - x^{2} }}$
i.e. $\tan(\sin^{-1}x)$ = $\frac{x}{\sqrt {1 - x^{2} }}$
