If y = tan^{-1}sqrt{frac{1 + cos x}{1 - cos x | vedclass

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

Question

(B) Given $y = 	an^{-1}\sqrt{\frac{1 + \cos x}{1 - \cos x}}$.Using the trigonometric identities $1 + \cos x = 2\cos^2(\frac{x}{2})$ and $1 - \cos x =

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

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

If $y = \tan^{-1}\sqrt{\frac{1 + \cos x}{1 - \cos x}}$,then $\frac{dy}{dx}$ is equal to

Similar Questions

Write the function in the simplest form: $\tan ^{-1}\left(\frac{1}{\sqrt{x^{2}-1}}\right), |x|>1$

If $\theta = \tan^{-1} a$,$\phi = \tan^{-1} b$ and $ab = -1$,then $\theta - \phi = $

If $\cot^{-1}[(\cos \alpha)^{1/2}] - \tan^{-1}[(\cos \alpha)^{1/2}] = x$,then $\sin x = $

If $\tan (x + y) = 33$ and $x = \tan^{-1}(3)$,then $y$ will be

If $\sin ^{-1}\left(\frac{x}{5}\right)+\operatorname{cosec}^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}$, then $5+x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module