Answer
$\displaystyle \frac{x}{\sqrt{1+x^{2}}}$
Work Step by Step
Let $t=\tan^{-1}x.$
Then, $t\in(- \displaystyle \frac{\pi}{2},\frac{\pi}{2})$ and $\tan t=x$
We can illustrate this in a right triangle where the leg opposite t is $x$
and the leg adjacent to $t$ is $1.$
(see below)
The hypotenuse is found by the Pythagorean theorem:
$\left[\begin{array}{l}
c^{2}=x^{2}+1^{2}\\
c=\sqrt{1+x^{2}}
\end{array}\right]$
So, $\displaystyle \sin t=\frac{opp}{hyp}=\frac{x}{\sqrt{1+x^{2}}}$
