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. $\frac{x}{\sqrt{1-x^2}}$
• D. $\frac{1}{\sqrt{1-x^2}}$

Answer 1

The correct option is A $\frac{x}{\sqrt{1+x^2}}$

To find: $\sin \left({\tan }^{-1}x\right)$
Let $\theta ={\tan }^{-1}x\text{so},x=\tan \theta$
$\sin \left({\tan }^{-1}x\right)=\sin \theta$
$\Rightarrow \sin \left({\tan }^{-1}x\right)=\frac{1}{\text{cosec}\theta }=\frac{1}{\sqrt{1+{\cot }^2\theta }}$
$\Rightarrow \sin \left({\tan }^{-1}x\right)=\frac{1}{\sqrt{1+{\cot }^2\theta }}$
$\Rightarrow \sin \left({\tan }^{-1}x\right)=\frac{1}{\sqrt{1+\frac{1}{{\tan }^2\theta }}}$
Now putting the values of $\theta$, we get
$\Rightarrow \sin \left({\tan }^{-1}x\right)=\frac{1}{\sqrt{1+\frac{1}{\left(\tan \right({\tan }^{-1}x){)}^2}}}$
$\Rightarrow \sin \left({\tan }^{-1}x\right)=\frac{1}{\sqrt{1+\frac{1}{x^2}}}=\frac{x}{\sqrt{x^2+1}}$
$\therefore$Option $\left(D\right)$ is correct.