Answer
$ \frac{x}{\sqrt{1 + x^2}}$
Work Step by Step
Let $\tan ^ {–1} x = \theta$
then, $\tan \theta = x$ and $\cot \theta = \frac{1}{x}$
To evaluate
$\sin (\tan ^ {–1} x) = \sin \theta$
We know that,
$\csc \theta = \sqrt{1 + \cot ^2 \theta}$
also $\sin \theta = \frac{1}{\csc \theta} $
Using above relations we get
$\csc \theta = \sqrt{1+\frac{1}{x^2}}$
$=> \csc \theta = \sqrt{\frac{1 + x^2}{x^2}}$
$=> \sin \theta = \frac{x}{\sqrt{1 + x^2}}$
