Application of Derivatives — Class 12 Maths Solution

We have, $f(x) = {\tan ^{ - 1}}(\sin x + \cos x)$

Therefore,${f^\prime }(x) = \frac{1}{{1 + {{(\sin x + \cos x)}^2}}} \cdot (\cos x - \sin x)$

$= \frac{1}{{1 + {{\sin }^2}x + {{\cos }^2}x + 2\sin x \cdot \cos x}}(\cos x - \sin x)$

$= \frac{1}{{(2 + \sin 2x)}}(\cos x - \sin x)$

[ and ${\sin ^2}x + {\cos ^2}x = 1$]

For ${f^\prime }(x) \ge 0$
$\frac{1}{{(2 + \sin 2x)}} \cdot (\cos x - \sin x) \ge 0$

$\Rightarrow$ $\cos x - \sin x \ge 0$

$\Rightarrow$ $\cos x \ge \sin x$

which is true, in $x \in \left( {0,\frac{\pi }{4}} \right)$.

Hence, $f(x)$ is an increasing function in $\left( {0,\frac{\pi }{4}} \right)$.