Given, the function, $f\left(x\right)={\tan }^{-1}(\sin x+\cos x)$ in $\left[0,\frac{\pi }{2}\right]$

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

$=\frac{\cos x-\sin x}{1+\sin 2x}$

$f^{\text{'}}\left(x\right)=0$

$\cos x-\sin x=0$

$\tan x=1$

$x=\frac{\mathrm{\pi }}{4}$
Then, $1\le \sin x+\cos x\le \sqrt{2} x\in \left[\frac{\mathrm{\pi }}{4},{\tan }^{-1}\sqrt{2}\right]$
$M={\tan }^{-1}\left(\sqrt{2}\right)$

$m={\tan }^{-1}\left(1\right)$

$M-m={\tan }^{-1}\left(\sqrt{2}\right)-{\tan }^{-1}\left(1\right)$

$M-m={\tan }^{-1}\left(\frac{\sqrt{2}-1}{1+\sqrt{2}}\right)$

$M-m={\tan }^{-1}(3-2\sqrt{2})$

$\tan (M-m)=3-2\sqrt{2}$