frac{d}{dx}left( tan^{-1} left( frac{cos x}{1 | vedclass

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Question

(A) Let $y = 	an^{-1} \left( \frac{\cos x}{1 + \sin x} ight)$.Using trigonometric identities $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$,$\s

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$ - \frac{1}{2}$

Vedclass · Answer

(A) Let $y = 	an^{-1} \left( \frac{\cos x}{1 + \sin x} ight)$.Using trigonometric identities $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$,$\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,and $1 = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}$:$y = 	an^{-1} \left( \frac{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}{(\cos \frac{x}{2} + \sin \frac{x}{2})^2} ight)$$y = 	an^{-1} \left( \frac{(\cos \frac{x}{2} - \sin \frac{x}{2})(\cos \frac{x}{2} + \sin \frac{x}{2})}{(\cos \frac{x}{2} + \sin \frac{x}{2})^2} ight)$$y = 	an^{-1} \left( \frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2}} ight)$Dividing numerator and denominator by $\cos \frac{x}{2}$:$y = 	an^{-1} \left( \frac{1 - 	an \frac{x}{2}}{1 + 	an \frac{x}{2}} ight) = 	an^{-1} \left( 	an \left( \frac{\pi}{4} - \frac{x}{2} ight) ight)$$y = \frac{\pi}{4} - \frac{x}{2}$Now,differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{4} - \frac{x}{2} ight) = - \frac{1}{2}$.

$\frac{d}{dx}\left( \tan^{-1} \left( \frac{\cos x}{1 + \sin x} \right) \right) = $

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