Question

$\sin \left(\tan ^{-1} x\right),|x|<1$ 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. $\frac{x}{\sqrt{1+x^2}}$

Answer 1

Answer: (D)

Let $\tan ^{-1} x=\theta$
$\Rightarrow x =\tan \theta$
$\Rightarrow \sec \theta =\sqrt{1+x^2} \left(\because \sec \theta=\sqrt{1+\tan ^2 \theta}\right)$
$\Rightarrow \cos \theta =\frac{1}{\sqrt{1+x^2}} $
$\therefore \sin \theta =\sqrt{1-\frac{1}{1+x^2}} \left(\because \sin \theta=\sqrt{1-\cos ^2 \theta}\right) $
$ =\sqrt{\frac{1+x^2-1}{1+x^2}}=\frac{x}{\sqrt{1+x^2}} $
$\therefore \sin \left(\tan ^{-1} x\right) =\sin \left(\sin ^{-1} \frac{x}{\sqrt{1+x^2}}\right)$
$=\frac{x}{\sqrt{1+x^2}}$