Answer
$\dfrac{x}{\sqrt {1+x^2}}$
Work Step by Step
Suppose $\theta =\tan^{-1} x $
This gives: $ x =\tan \theta $
Since, $ r=\sqrt {a^2+y^2}$
This implies that $ r=\sqrt {(1)^2+x^2}=\sqrt {1+x^2}$
Therefore, we have $\sin(\tan^{-1} x)=\dfrac{x}{\sqrt {1+x^2}}$
