$\sin ({\tan ^{ - 1}}x),\;|x|\; < 1$ is equal to

(A) $\frac{x}{{\sqrt {1 - {x^2}} }}$

(B) $\frac{1}{{\sqrt {1 - {x^2}} }}$

(C) $\frac{1}{{\sqrt {1 + {x^2}} }}$

(D) $\frac{x}{{\sqrt {1 + {x^2}} }}$

Option D is correct

Let ${\tan ^{ - 1}}x = \theta \Rightarrow x = \tan \theta ,\;\;where\;\;\theta \in \left( { - \frac{\pi }{2},\;\frac{\pi }{2}} \right)$
$\therefore$ $\sin ({\tan ^{ - 1}}x) = \sin \theta$

Now, $\sin \theta = \frac{1}{{{\rm{cosec }}\theta }} = \frac{1}{{\sqrt {1 + {{\cot }^2}\theta } }} = \frac{1}{{\sqrt {1 + \frac{1}{{{{\tan }^2}\theta }}} }} = \frac{x}{{\sqrt {{x^2} + 1} }}$