Official solution
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\[({\tan ^{ - 1}}\sqrt 2 - {\pi \over 4}) = {{\sqrt 2 - 1} \over {1 + \sqrt 2 }} \times {{\sqrt 2 - 1} \over {\sqrt 2 - 1}} = 3 - 2\sqrt 2 \]