Answer
$sin (tan^{-1}~x) = \frac{x}{\sqrt{x^2+1}}$
Work Step by Step
Suppose that: $tan ~\theta = x$
Then: $\frac{opp}{adj} = \frac{x}{1}$
Then: $hyp = \sqrt{x^2+1}$
We can find $sin (tan^{-1}~x)$:
$sin (tan^{-1}~x)$
$= sin (\theta)$
$= \frac{opp}{hyp}$
$= \frac{x}{\sqrt{x^2+1}}$
