$f(x) = {\tan ^{ - 1}}(\sin x - \cos x),\,\,\,\,\,[0,\pi ]$ 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 ${f_{\min }} = {\tan ^{ - 1}}\sqrt 2 - {\pi \over 4}$ $= {\cos ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right) - {\pi \over 4}$