Explanation

(b)
Since, $f(x) = \tan^{- 1}{(\sin{x + \cos{x)}}}$
$\Rightarrow f^{'}(x) = \frac{1}{1 + (\sin{x + \cos{x\left. \ \ \right)^{2}\ }}}(\cos{x - \sin{x)}}$
$= \frac{\sqrt{2}\cos\left( x + \frac{\pi}{4} \right)}{1 + (\sin{x + \cos{x\left. \ \ \right)^{2}}}}$
For $f(x)$ to be increasing,
$- \frac{\pi}{2} < x\frac{\pi}{4} < \frac{\pi}{2}$
$\Rightarrow - \frac{3\pi}{4} < x < \frac{\pi}{4}$
Hence, option (b) is correct, which lies in the above interval.