$The function$ $f (x) = \tan^{- 1} (sinx + cosx) is an$ $increasing function in$

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$(\frac{π}{4}, \frac{π}{2})$

$(0, \frac{π}{2})$

$(- \frac{π}{2}, \frac{π}{4})$

$(- \frac{π}{2}, \frac{π}{2})$

answer is D.

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Detailed Solution

$\begin{array}{l} Given f (x) = \tan^{- 1} (\sin x + \cos x) \\ Differentiate both sides with respect to x \\ f^{'} (x) = \frac{1}{1 + {(\sin x + \cos x)}^{2}} (\cos x - \sin x) \\ = \frac{\sqrt{2}}{1 + {(\sin x + \cos x)}^{2}} (\frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x) \\ = \frac{\sqrt{2}}{1 + {(\sin x + \cos x)}^{2}} (\cos \frac{π}{4} \cos x - \sin \frac{π}{4} \sin x) \\ = \frac{\sqrt{2} \cos (x + \frac{π}{4})}{1 + {(\sin x + \cos x)}^{2}} \\ If f^{'} (x) > 0 then f (x) is increasing function \\ hence f (x) is increasing if - \frac{π}{2} < x + \frac{π}{4} < \frac{π}{2} \\ It gives x \in [- \frac{3 π}{4}, \frac{π}{4}] \end{array}$