Question:
Solve $\sin \left(\tan ^{-1} x\right),|x|<1$ 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) $\frac{x}{\sqrt{1+x^{2}}}$
Solution:
Let $\tan ^{-1} x=y$. Then, $\tan y=x \Rightarrow \sin y=\frac{x}{\sqrt{1+x^{2}}}$.
$\therefore y=\sin ^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right) \Rightarrow \tan ^{-1} x=\sin ^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right)$
$\therefore \sin \left(\tan ^{-1} x\right)=\sin \left(\sin ^{-1} \frac{x}{\sqrt{1+x^{2}}}\right)=\frac{x}{\sqrt{1+x^{2}}}$
The correct answer is $D$.