sin (tan^{-1} x),where |x|

Question

(A) Let $	an^{-1} x = 	heta$. Then,$	an 	heta = x = \frac{x}{1}$. In a right-angled triangle,the opposite side is $x$ and the adjacent side is $1$

Vedclass · Accepted Answer

$\frac{x}{\sqrt{1+x^2}}$

Vedclass · Answer

(A) Let $	an^{-1} x = 	heta$. Then,$	an 	heta = x = \frac{x}{1}$. In a right-angled triangle,the opposite side is $x$ and the adjacent side is $1$. The hypotenuse is $\sqrt{(	ext{opposite})^2 + (	ext{adjacent})^2} = \sqrt{x^2 + 1^2} = \sqrt{1+x^2}$. Now,$\sin 	heta = \frac{	ext{opposite}}{	ext{hypotenuse}} = \frac{x}{\sqrt{1+x^2}}$. Therefore,$\sin (	an^{-1} x) = \frac{x}{\sqrt{1+x^2}}$. The correct option is $A$.

$\sin (\tan^{-1} x)$,where $|x| < 1$,is equal to:

Similar Questions

The value of $\sin^{-1}\left(\cos \frac{53\pi}{5}\right)$ is

$\tan^{-1} \frac{x}{\pi} < \frac{\pi}{3}, x \in N$,then the maximum value of $x$ is :-

$\tan \left( 2{{\cos }^{ - 1}}\frac{3}{5} \right) = $

$\tanh ^{-1}\left(\frac{1}{2}\right)+\operatorname{coth}^{-1}(2)$ is equal to

$\cos \left[\cot ^{-1}(-\sqrt{3})+\frac{\pi}{6}\right]$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module