#### Question

Show that f(x) = tan^{−1} (sin x + cos x) is a decreasing function on the interval (π/4, π/2) ?

#### Solution

\[f\left( x \right) = \tan^{- 1} \left( \sin x + \cos x \right)\]

\[f'\left( x \right) = \frac{1}{1 + \left( \sin x + \cos x \right)^2}\left( \cos x - \sin x \right)\]

\[ = \frac{1}{1 + 1 + 2 \sin x \cos x}\left( \cos x - \sin x \right)\]

\[ = \frac{\left( \cos x - \sin x \right)}{2 + \sin 2x}\]

\[\text { Here },\]

\[\frac{\pi}{4} < x < \frac{\pi}{2}\]

\[ \Rightarrow \frac{\pi}{2} < 2x < \pi\]

\[ \Rightarrow \sin 2x > 0\]

\[ \Rightarrow 2 + \sin 2x > 0 . . . \left( 1 \right)\]

\[\text { Also,} \]

\[\frac{\pi}{4} < x < \frac{\pi}{2}\]

\[\cos x < \sin x\]

\[ \Rightarrow \cos x - \sin x < 0 . . . \left( 2 \right)\]

\[f'\left( x \right) = \frac{\left( \cos x - \sin x \right)}{2 + \sin 2x} < 0, \forall x \in \left( \frac{\pi}{4}, \frac{\pi}{2} \right) \left[ \text { From eqs } . (1) \text { and } (2) \right]\]

\[\text { So },f\left( x \right)\text { is decreasing on }\left( \frac{\pi}{4}, \frac{\pi}{2} \right).\]