Let M and m respectively be the maximum and minimum values of the...

Given, the function, $f ( x ) = \tan^{- 1} ( \sin x + \cos x )$ in \left[0, \dfrac{\pi}{2}\right]$</p><p>${f}^{'}\left(x\right) = \dfrac{\cos x - \sin x}{1 + {\left(\sin x + \cos x\right)}^{2}}$</p><p>$= \dfrac{\cos x - \sin x}{1 + \sin 2x}$</p><p>${f}^{'}\left(x\right) = 0$</p><p>$\cos x - \sin x = 0$</p><p>$\tan x = 1$</p><p>$x = \dfrac{\pi}{4}$<br />Then,$1 \le \sin x + \cos x \le \sqrt{2}x \in \left[\dfrac{\pi}{4}, \tan^{- 1}\sqrt{2}\right]$<br />$M = \tan^{- 1} ( \sqrt{2} )$</p><p>$m = \tan^{- 1} ( 1 )$</p><p>$M - m = \tan^{- 1}\left(\sqrt{2}\right) - $\tan^{- 1}\left(1\right)$

$M - m = \tan^{- 1}\left(\dfrac{\sqrt{2} - 1}{1 + \sqrt{2}}\right)</p><p>$M - m = \tan^{- 1} ( 3 - 2\sqrt{2} )$</p><p>$\tan ( M - m ) = 3 - 2\sqrt{2}$</p>$