Prove that $\tan^{- 1} (\frac{\cos x}{1 + \sin x}) = \frac{π}{4} - \frac{π}{2}, x \in (- \frac{π}{2}, \frac{π}{2})$

Solution

$\tan^{- 1} (\frac{\cos x}{1 + \sin x}) = \tan^{- 1} [\frac{\sin (\frac{π}{2} - x)}{1 + \cos (\frac{π}{2} - x)}] = \tan^{- 1} [\frac{2 \sin (\frac{π}{4} - \frac{x}{2}) \cos (\frac{π}{4} - \frac{x}{2})}{2 \cos^{2} (\frac{π}{4} - \frac{x}{2})}] . . . . . . . . . . . . . . [∵ \sin θ = 2 \sin (θ / 2) \cos (θ / 2) and 1 + \cos θ = 2 \cos^{2} (θ / 2)]$

$= \tan^{- 1} [\tan (\frac{π}{4} - \frac{π}{2})] = (\frac{π}{4} - \frac{π}{2})$

Hence proved.

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Prove that $\tan^{- 1} (\frac{\cos x}{1 + \sin x}) = \frac{π}{4} - \frac{π}{2}, x \in (- \frac{π}{2}, \frac{π}{2})$

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Prove that $\tan^{- 1} (\frac{\cos x}{1 + \sin x}) = \frac{π}{4} - \frac{π}{2}, x \in (- \frac{π}{2}, \frac{π}{2})$