If $y = an^{- 1}\left( \frac{\cos x}{1 + \sin x} ight)$, then $\frac{dy}{dx}$ is equal to

Question

If $y = 	an^{- 1}\left( \frac{\cos x}{1 + \sin x} ight)$, then $\frac{dy}{dx}$ is equal to

Dr. Abubacker · Accepted Answer

(d)
$\frac{\cos x}{1 + \sin x} = \frac{\cos^{2}{\frac{x}{2} - \sin^{2}\frac{x}{2}}}{\left( \cos{\frac{x}{2} + \sin\frac{x}{2}} ight)^{2}}$
$= \frac{\cos{\frac{x}{2} - \sin\frac{x}{2}}}{\cos{\frac{x}{2} + \sin\frac{x}{2}}}\ \ \ \  = \frac{1 - 	an\frac{x}{2}}{1 + 	an\frac{x}{2}}$
=$\ 	an\left( \frac{\pi}{4} - \frac{x}{2} ight)$
$	herefore = 	an^{- 1}{\left( 	an\left( \frac{\pi}{4} - \frac{x}{2} ight) ight) = \frac{\pi}{4} - \frac{x}{2}\ }$
$\Rightarrow \ \ \ \ \ \ \ \frac{dy}{dx} = - \frac{1}{2}$

Q. If $y = \tan^{- 1}\left( \frac{\cos x}{1 + \sin x} \right)$, then $\frac{dy}{dx}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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