The sum  of absolute maximum and absolute minimum values of the function
$f\left(x\right)={\tan }^{-1}(\sin x-\cos x)$  in  $[0,\pi ]$   is 
 

$$\begin{gathered} f\left(x\right)={\tan }^{-1}(\sin x-\cos x) f^1\left(x\right)=\frac{\cos x+\sin x}{1+{(\sin x-\cos x)}^2} f^1\left(x\right)=0\Rightarrow \tan x=-1 \\ \Rightarrow x=\frac{3\pi }{4}\in (0,\pi ) f\left(0\right)=\frac{-\pi }{4} f\left(\pi \right)=\frac{\pi }{4} f\left(\frac{3\pi }{4}\right)={\tan }^{-1}\left(\sqrt{2}\right) \\ Absolute max + absolute min= {\tan }^{-1}\left(\sqrt{2}\right)-\frac{\pi }{4} \end{gathered}$$