Let g(x) = sin x + cos x = $\sqrt 2$ sin$\left( {x + {\pi \over 4}} \right)$ g(x)$\in$ $\left[ {1,\sqrt 2 } \right]$ for x$\in$ [0, $\pi$/2] f(x) = tan $-$1 (sin x + cos x) $\in$ $\left[ {{\pi \over 4},{{\tan }^{ - 1}}\sqrt 2 } \right]$ tan$({\tan ^{ - 1}}\sqrt 2 - {\pi \over 4}) = {{\sqrt 2 - 1} \over {1 + \sqrt 2 }} \times {{\sqrt 2 - 1} \over {\sqrt 2 - 1}} = 3 - 2\sqrt 2$