The sum of the absolute maximum and absolute minimum values of the function

$f\left(x\right)={\tan }^{-1}\left(\sin x- \cos x\right) ind the interval \left[0, \pi \right] is$

$$\begin{gathered} f\left(x\right)={\tan }^{-1}\left(\sin x-\cos x\right) \\ \Rightarrow f^{\text{'}}\left(x\right)=\frac{\cos x+\sin x}{{\left(\sin x-\cos x\right)}^2+1}=0 \\ \therefore x=\frac{3\pi }{4} \end{gathered}$$

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

$\begin{array}{rcl}\mathrm{sum} \mathrm{of} \mathrm{maximum} \mathrm{and} \mathrm{minimum}& =& {\tan }^{-1}\sqrt{2}-\frac{\mathrm{\pi }}{4}\\ & =& {\cos }^{-1}\frac{1}{\sqrt{3}}-\frac{\mathrm{\pi }}{4}\end{array}$