Let the function $f(x)=\tan ^{-1}(\sin x+\cos x)$ be defined on $[0,2 \pi]$. Then, f(x) is

increasing on $[0, \pi / 4) \cup(5 \pi / 4,2 \pi]$

We have,

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

$\Rightarrow f'(x) =\frac{1}{1+(\sin x+\cos x)^2} \times(\cos x-\sin x)$

Now,

$f'(x)>0 \Rightarrow \cos x-\sin x>0 \Rightarrow x \in[0, \pi / 4) \cup(5 \pi / 4,2 \pi]$

and, $f'(x)<0 \Rightarrow \cos x-\sin x<0 \Rightarrow x \in(\pi / 4,5 \pi / 4)$

Hence, f(x) is increasing on $[0, \pi / 4) \cup(5 \pi / 4,2 \pi]$ and decreasing on $(\pi / 4,5 \pi / 4)$.