frac{d}{dx}left( tan^{-1}sqrt{frac{1 + cos(x/ | vedclass

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

Question

(A) Let $y = 	an^{-1}\sqrt{\frac{1 + \cos(x/2)}{1 - \cos(x/2)}}$.Using the trigonometric identities $1 + \cos 	heta = 2\cos^2(	heta/2)$ and $1 - \c

Vedclass · Accepted Answer

$- \frac{1}{4}$

Vedclass · Answer

(A) Let $y = 	an^{-1}\sqrt{\frac{1 + \cos(x/2)}{1 - \cos(x/2)}}$.Using the trigonometric identities $1 + \cos 	heta = 2\cos^2(	heta/2)$ and $1 - \cos 	heta = 2\sin^2(	heta/2)$,we get:$y = 	an^{-1}\sqrt{\frac{2\cos^2(x/4)}{2\sin^2(x/4)}}$$y = 	an^{-1}\sqrt{\cot^2(x/4)}$$y = 	an^{-1}(\cot(x/4))$Since $\cot(x/4) = 	an(\frac{\pi}{2} - \frac{x}{4})$,we have:$y = 	an^{-1}(	an(\frac{\pi}{2} - \frac{x}{4})) = \frac{\pi}{2} - \frac{x}{4}$Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{2} - \frac{x}{4}) = 0 - \frac{1}{4} = - \frac{1}{4}$.

$\frac{d}{dx}\left( \tan^{-1}\sqrt{\frac{1 + \cos(x/2)}{1 - \cos(x/2)}} \right)$ is equal to

Similar Questions

If $f(x)=e^x$,$g(x)=\sin^{-1} x$ and $h(x)=f(g(x))$,then $\frac{h^{\prime}(x)}{h(x)}$ is

$\frac{d}{{dy}}\left( {{{\sin }^{ - 1}}\left( {\frac{{3y}}{2} - \frac{{{y^3}}}{2}} \right)} \right) = $

$\frac{d}{dx} \sin^{-1}(2ax\sqrt{1 - a^2x^2}) = $

$\frac{d}{dx} \tan^{-1} \left( \frac{4\sqrt{x}}{1 - 4x} \right) = $

If $2y = {\left( {{{\cot }^{ - 1}}\left( {\frac{{\sqrt 3 \cos x + \sin x}}{{\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}$ and $x \in \left( {0,\frac{\pi }{2}} \right)$,then $\frac{{dy}}{{dx}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module