Answer
$\frac{x}{\sqrt {1 + x^{2}}}$
Work Step by Step
Given expression is-
$\sin(\tan^{-1}x)$
Assuming $\tan^{-1} x$ = $u$ , we get-
$\tan u$ = $x$, [$-\frac{\pi}{2}\leq u\leq \frac{\pi}{2}$]
Considering right triangle ABC in which, $\tan u$ = $x$
i.e. in right triangle ABC, $\frac{opp}{adj}$ = $\frac{x}{1}$
From Pythagorean Theorem we know
$hypo$ = $\sqrt {adj^{2} +opp^{2} }$
i.e. $hypo$ = $\sqrt {1^{2} + x^{2}}$ = $\sqrt {1 + x^{2}}$
Therefore-
$\sin u$ = $\frac{opp}{hypo}$
i.e. $\sin u$ = $\frac{x}{\sqrt {1 + x^{2}}}$
i.e. $\sin(\tan^{-1}x)$ = $\frac{x}{\sqrt {1 + x^{2}}}$
