# Show that $f(x)=\tan^{-1}(\sin x+\cos x)$ is an increasing function in$\bigg(0,\Large \frac{1}{4}\normalsize \bigg).$

Toolbox:
• Let $f(x)$ be a function defined on $(a,b)$. If $f'(x) > 0$ for all $x \in (a,b)$ except for a finite number of points, where $f'(x)>0,$ then $f(x)$ is increasing on $(a,b)$
Step 1
$f(x)= \tan^{-1} ( \sin x+ \cos x)$
Differentiating w.r.t $x$ we get,
$f'(x)=\large\frac{1}{1+ ( \sin x+ \cos x)^2} \times ( \cos x- \sin x)$
$= \large\frac{\cos x- \sin x}{1+( \sin x+ \cos x)^2}= \large\frac{\cos x(1- \tan x)}{1+ ( \sin x+ \cos x)^2}$
Now let us consider the interval $0 < x < \large\frac{\pi}{4}$
$\Rightarrow \cos x > 0, \large\frac{1}{1+(\sin x+ \cos x)^2}\: and \: (1- \tan x) > 0$
$( \because \tan x < 1$ in the interval $0 < x < \large\frac{\pi}{4})$
$\Rightarrow \large\frac{\cos x(1-\tan x)}{1+ ( \sin x+ \cos x)^2} > 0$
$\Rightarrow f'(x) > 0$
Hence $f'(x)>0$ for all $x \in \bigg( 0, \large\frac{\pi}{4} \bigg)$
Hence $f(x)$ is increasing on $\bigg( 0, \large\frac{\pi}{4} \bigg)$
answered Aug 5, 2013