The sum of the absolute maximum and absolute minimum values of the...

Given,

$f\left(x\right) = \tan^{- 1}\left(\sin x - \cos x\right)$

Now differentiating the function we get,

${f}^{'}\left(x\right) = \dfrac{\cos x + \sin x}{{\left(\sin x - \cos x\right)}^{2} + 1}$

Now equating ${f}^{'}\left(x\right) = 0$ to find critical points,

$\dfrac{\cos x + \sin x}{{\left(\sin x - \cos x\right)}^{2} + 1} = 0$

$⇒ \cos x + \sin x = 0$

$⇒ \tan x = - 1$

$∴ x = \dfrac{3\pi}{4}$

Now checking the value of function at critical point and boundary point we get,

$x$	$0$	$\dfrac{3\pi}{4}$	$\pi$
$f\left(x\right)$	$- \dfrac{\pi}{4}$	$\tan^{- 1}\sqrt{2}$	$\dfrac{\pi}{4}$

So, we get,

\left\begin{array}{ll}{\left(f\left(x\right)\right)}_\max = \tan^{- 1}\sqrt{2} \\ ∴ {\left(f\left(x\right)\right)}_\min = - \dfrac{\pi}{4}\end{array}\right]$</p><p>So, the sum will be$= \tan^{- 1}\sqrt{2} - \dfrac{\pi}{4}$

= \cos^{- 1}\dfrac{1}{\sqrt{3}} - \dfrac{\pi}{4}