$$f(x) = {\tan ^{ - 1}}(\sin x - \cos x),\,\,\,\,\,[0,\pi ]$$

Let $$g(x) = \sin x - \cos x$$

$$ = \sqrt 2 \sin \left( {x - {\pi \over 4}} \right)$$ and $$x - {\pi \over 4} \in \left[ {{{ - \pi } \over 4},\,{{3\pi } \over 4}} \right]$$

$$\therefore$$ $$g(x) \in \left[ { - 1,\,\sqrt 2 } \right]$$

and $${\tan ^{ - 1}}x$$ is an increasing function

$$\therefore$$ $$f(x) \in \left[ {{{\tan }^{ - 1}}( - 1),\,{{\tan }^{ - 1}}\sqrt 2 } \right]$$

$$ \in \left[ { - {\pi \over 4},\,{{\tan }^{ - 1}}\sqrt 2 } \right]$$

$$\therefore$$ Sum of $${f_{\max }}$$ and $${f_{\min }} = {\tan ^{ - 1}}\sqrt 2 - {\pi \over 4}$$

$$ = {\cos ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right) - {\pi \over 4}$$