$$\tan^{-1}\left(\frac{\cos(x)-\sin(x)}{\cos(x)+\sin(x)}\right),\quad 0<x<\pi$$
$=\tan^{-1}\left(\frac{1-\tan(x)}{1+\tan(x)}\right)$ diving the numerator and denominator by $\cos x$
$=\tan^{-1}\left(\frac{\tan\frac{\pi}{4}-\tan(x)}{1+\tan\frac{\pi}{4}\tan(x)}\right)$ as $\tan\frac{\pi}{4}=1$
$=\tan^{-1}\tan(\frac{\pi}{4}-x)$
$=n\pi+\frac{\pi}{4}-x$ where n is any integer.
The principal value which must lie in $[-\frac{\pi}{2},\frac{\pi}{2}]$,
will be
$\frac{\pi}{4}-x$ when $\frac{\pi}{4}-x$ lies in that region i.e, $\frac{3\pi}{4} ≥ x > -\frac{\pi}{4}$
and $\pi +\frac{\pi}{4}-x$ elsewhere.
But $0<x<\pi$,
so, if $\frac{\pi}{2} ≥ x ≥ 0$ the principal value =$\frac{\pi}{4}-x$
and $\frac{5\pi}{4}-x$ elsewhere.