Answer
$sin(arctan~x)=±\frac{x}{\sqrt {x^2+1}}$
Work Step by Step
Let $u=arctan~x~~$ (Range: $-\frac{\pi}{2}\lt u\lt\frac{\pi}{2}$).
Then: $x=tan~u$
$cot~u=\frac{1}{tan~u}=\frac{1}{x}$
$csc^2u=1+cot^2u$
$csc^2u=1+(\frac{1}{x})^2$
$csc^2u=1+\frac{1}{x^2}=\frac{x^2+1}{x^2}$
$sin^2u=\frac{1}{csc^2u}=\frac{x^2}{x^2+1}$
$sin~u=±\frac{x}{\sqrt {x^2+1}}$
