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