frac{d}{d x} tan ^{-1}left[frac{sqrt{1+sin x} | vedclass

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

Question

(C) Let $y = 	an ^{-1}\left[\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}ight]$.We know that $1 + \sin x = \cos^2 \frac{x

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(C) Let $y = 	an ^{-1}\left[\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}ight]$.We know that $1 + \sin x = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$.Similarly, $1 - \sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$.Substituting these into the expression:$y = 	an ^{-1}\left[\frac{(\cos \frac{x}{2} + \sin \frac{x}{2}) - (\cos \frac{x}{2} - \sin \frac{x}{2})}{(\cos \frac{x}{2} + \sin \frac{x}{2}) + (\cos \frac{x}{2} - \sin \frac{x}{2})}ight]$$y = 	an ^{-1}\left[\frac{2 \sin \frac{x}{2}}{2 \cos \frac{x}{2}}ight] = 	an ^{-1}(	an \frac{x}{2}) = \frac{x}{2}$.Now, differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\frac{x}{2}) = \frac{1}{2}$.

$\frac{d}{d x} \tan ^{-1}\left[\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}\right]$ is equal to

Similar Questions

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

The differential coefficient of ${\sec ^{ - 1}}\left( \frac{1}{{2{x^2} - 1}} \right)$ with respect to $\sqrt {1 - {x^2}} $ at $x = \frac{1}{2}$ is:

For $-1 < x < 1$,if $f(x) = \cos^2 \left( \tan^{-1} \sqrt{\frac{1-x}{1+x}} \right)$,then $f'(x) =$

$\frac{d}{dx} \left\{ \sin^2 \left( \cot^{-1} \sqrt{\frac{1 + x}{1 - x}} \right) \right\} =$

Differentiate the following with respect to $x$: $\sin ^{-1}\left(\frac{2^{x+1}}{1+4^{x}}\right)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module