sin [cot ^{ - 1}(cos tan ^{ - 1}x)] = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $	an^{-1}x = 	heta$,then $	an 	heta = x$. We know that $\cos(	an^{-1}x) = \cos 	heta = \frac{1}{\sqrt{1+x^2}}$. Now,the expression becom

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $	an^{-1}x = 	heta$,then $	an 	heta = x$. We know that $\cos(	an^{-1}x) = \cos 	heta = \frac{1}{\sqrt{1+x^2}}$. Now,the expression becomes $\sin[\cot^{-1}(\frac{1}{\sqrt{1+x^2}})]$. Let $\cot^{-1}(\frac{1}{\sqrt{1+x^2}}) = \phi$,then $\cot \phi = \frac{1}{\sqrt{1+x^2}}$. This implies $	an \phi = \sqrt{1+x^2}$. Using the identity $\sin \phi = \frac{	an \phi}{\sqrt{1+	an^2 \phi}}$,we get: $\sin \phi = \frac{\sqrt{1+x^2}}{\sqrt{1+(1+x^2)}} = \sqrt{\frac{1+x^2}{2+x^2}}$. Thus,the correct option is $D$.

$\sin [\cot ^{ - 1}(\cos \tan ^{ - 1}x)] =$

Similar Questions

If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$,then the value of $x^2+y^2+z^2-2xyz$ is

The value of ${\tan ^{ - 1}}\left[ {\cos \left( {2\,{{\tan }^{ - 1}}\frac{3}{4}} \right)\, + \,\sin \,\left( {2\,{{\cot }^{ - 1}}\frac{1}{2}} \right)} \right]$ is

If $\angle A = 90^\circ$ in the triangle $ABC$,then $\tan^{-1}\left(\frac{c}{a+b}\right) + \tan^{-1}\left(\frac{b}{a+c}\right) = $

If $\cos^{-1} x + \cos^{-1} y = 2\pi$,then $\sin^{-1} x + \sin^{-1} y$ is equal to

If $\frac{1}{2} \leq x \leq 1$, then $\cos ^{-1} x+\cos ^{-1}\left(\frac{x}{2}+\frac{1}{2} \sqrt{3-3 x^2}\right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module