Answer
$\sin ^{}\left( \tan^{-1} x\right) =\dfrac {x}{\sqrt {1+x^{2}}}$
Work Step by Step
Lets assume $\alpha =\tan ^{-1}x\Rightarrow \tan \alpha =x$
$\sin \left( \tan ^{-1}x\right) =\sin \alpha $
$\tan \alpha =\dfrac {\sin \alpha }{\cos \alpha }=\dfrac {\sin \alpha }{\sqrt {1-\sin ^{2}\alpha }}=x$
$\Rightarrow \dfrac {\sin ^{2}\alpha }{1-\sin ^{2}\alpha }=x^{2}\Rightarrow \sin ^{2}\alpha =x^{2}-x^{2}\sin ^{2}\alpha \Rightarrow \sin ^{2}\alpha \left( 1+x^{2}\right) =x^{2}\Rightarrow \sin \alpha =\dfrac {x}{\sqrt {1+x^{2}}}$
$\Rightarrow \sin \alpha =\sin ^{}\left( \tan^{-1} x\right) =\dfrac {x}{\sqrt {1+x^{2}}}$
