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

Given,

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

Now differentiating the function we get,

$f^{\text{'}}\left(x\right)=\frac{\cos x+\sin x}{{\left(\sin x-\cos x\right)}^2+1}$

Now equating $f^{\text{'}}\left(x\right)=0$ to find critical points,

$\frac{\cos x+\sin x}{{\left(\sin x-\cos x\right)}^2+1}=0$

$\Rightarrow \cos x+\sin x=0$

$\Rightarrow \tan x=-1$

$\therefore x=\frac{3\pi }{4}$

Now checking the value of function at critical point and boundary point we get,

$x$	$0$	$\frac{3\pi }{4}$	$\pi$
$f\left(x\right)$	$-\frac{\pi }{4}$	${\tan }^{-1}\sqrt{2}$	$\frac{\pi }{4}$

So, we get,

$\left.\begin{array}{c}{\left(f\left(x\right)\right)}_{\max }={\tan }^{-1}\sqrt{2}\\ \therefore {\left(f\left(x\right)\right)}_{\min }=-\frac{\pi }{4}\end{array}\right]$

So, the sum will be $={\tan }^{-1}\sqrt{2}-\frac{\pi }{4}$

$={\cos }^{-1}\frac{1}{\sqrt{3}}-\frac{\pi }{4}$