For $0

Question

For $0 < x < 2,\frac{d}{dx}\ \left( 	an^{- 1}\sqrt{\frac{1 + \cos\frac{x}{2}}{1 - \cos\frac{x}{2}}} ight)\$is equal to

Dr. Abubacker · Accepted Answer

(a)
$\frac{d}{dx} = \left( 	an^{- 1}\sqrt{\frac{1 + \cos\frac{x}{2}}{1 - \cos\frac{x}{2}}} ight)\  = \ \frac{d}{dx}	an^{- 1}\sqrt{\frac{2\cos^{2}\frac{x}{4}}{2\sin^{2}\frac{x}{4}}}$
$= \frac{d}{dx}\ 	an^{- 1}\left| \cot\left( \frac{x}{4} ight) ight|$
$= \frac{d}{dx}	an^{- 1}\left| 	an\left( \frac{\pi}{2} - \frac{x}{4} ight) ight|$
$= \frac{d}{dx}\left( \frac{\pi}{2} - \frac{x}{4} ight) = \frac{- 1}{4}$

Q. For $0 < x < 2,\frac{d}{dx}\ \left( \tan^{- 1}\sqrt{\frac{1 + \cos\frac{x}{2}}{1 - \cos\frac{x}{2}}} \right)\$is equal to

Explanation

Reviewed by Dr. Abubacker

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