"The sum of the absolute maximum and absolute minimum values of the function $f(x)=\tan ^{-1}(\sin x-\cos x)$ in the interval $[0, \pi]$ is :"

Question

Accepted Answer

"

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

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 ${f_{\min }} = {\tan ^{ - 1}}\sqrt 2 - {\pi \over 4}$

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

"

The sum of the absolute maximum and absolute minimum values of the function $f(x)=\tan ^{-1}(\sin x-\cos x)$ in the interval $[0, \pi]$ is :