Solve
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 $D$.

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