If y = tan^{-1} left[ sqrt{frac{1 + cos(x/2)} | vedclass

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

Question

(B) Given $y = 	an^{-1} \left[ \sqrt{\frac{1 + \cos(x/2)}{1 - \cos(x/2)}} ight]$.Using the trigonometric identities $1 + \cos 	heta = 2 \cos^2(	h

Vedclass · Accepted Answer

$\frac{-1}{4}$

Vedclass · Answer

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

If $y = \tan^{-1} \left[ \sqrt{\frac{1 + \cos(x/2)}{1 - \cos(x/2)}} \right]$,then $\frac{dy}{dx} = $

Similar Questions

The value of $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) - \sin^{-1}\left(\frac{1}{2}\right)$ is ....... $^o$.

If ${\cos ^{ - 1}}\left( {\frac{1}{x}} \right) = \theta $,then $\tan \theta =$

The value of $\sin^{-1}(\sin 10)$ is

$A$ value of $x$ for which $\sin(\cot^{-1}(1 + x)) = \cos(\tan^{-1}x)$ is

$\sin (\cot ^{ - 1}x) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module