Evaluate: {tan ^{ - 1}}left[ {frac{{cos x}}{{ | vedclass

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Question

(A) We know that $\cos x = \sin(\frac{\pi}{2} - x)$ and $1 + \sin x = 1 + \cos(\frac{\pi}{2} - x)$.Using the trigonometric identities $\sin 	heta = 2

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) We know that $\cos x = \sin(\frac{\pi}{2} - x)$ and $1 + \sin x = 1 + \cos(\frac{\pi}{2} - x)$.Using the trigonometric identities $\sin 	heta = 2\sin(\frac{	heta}{2})\cos(\frac{	heta}{2})$ and $1 + \cos 	heta = 2\cos^2(\frac{	heta}{2})$:${	an ^{ - 1}}\left[ {\frac{{\cos x}}{{1 + \sin x}}} ight] = {	an ^{ - 1}}\left[ {\frac{{\sin (\pi /2 - x)}}{{1 + \cos (\pi /2 - x)}}} ight]$$= {	an ^{ - 1}}\left[ {\frac{{2\sin (\pi /4 - x/2)\cos (\pi /4 - x/2)}}{{2\cos^2 (\pi /4 - x/2)}}} ight]$$= {	an ^{ - 1}}\left[ {	an (\pi /4 - x/2)} ight]$$= \frac{\pi }{4} - \frac{x}{2}$.

Evaluate: ${\tan ^{ - 1}}\left[ {\frac{{\cos x}}{{1 + \sin x}}} \right]$

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