Express tan ^{-1} left( frac{cos x}{1-sin x} | vedclass

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Question

(D) Given expression: $	an ^{-1} \left( \frac{\cos x}{1-\sin x} ight)$Using trigonometric identities $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}

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$\frac{\pi}{4} + \frac{x}{2}$

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(D) Given expression: $	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 $1 = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}$:$= 	an ^{-1} \left[ \frac{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2}} ight]$$= 	an ^{-1} \left[ \frac{(\cos \frac{x}{2} - \sin \frac{x}{2})(\cos \frac{x}{2} + \sin \frac{x}{2})}{(\cos \frac{x}{2} - \sin \frac{x}{2})^2} ight]$$= 	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 $\cos \frac{x}{2}$:$= 	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]$$= \frac{\pi}{4} + \frac{x}{2}$

Express $\tan ^{-1} \left( \frac{\cos x}{1-\sin x} \right)$,$-\frac{3 \pi}{2} < x < \frac{\pi}{2}$ in the simplest form.

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