${\tan ^{ - 1}}\left( {\frac{{a\cos x - b\sin x}}{{b\cos x + a\sin x}}} \right),\frac{{ - \pi }}{2} < x < \frac{\pi }{2}$ and $\frac{a}{b}\tan x > - 1$

Let $y = {\tan ^{ - 1}}\left( {\frac{{a\cos x - b\sin x}}{{b\cos x + a\sin x}}} \right)$

$= {\tan ^{ - 1}}\left[ {\frac{{\frac{{a\cos x}}{{b\cos x}} - \frac{{b\sin x}}{{b\cos x}}}}{{\frac{{b\cos x}}{{b\cos x}} + \frac{{a\sin x}}{{b\cos x}}}}} \right] = {\tan ^{ - 1}}\left[ {\frac{{\frac{a}{b} - \tan x}}{{1 + \frac{a}{b}\tan x}}} \right]$

$= {\tan ^{ - 1}}\frac{a}{b} - {\tan ^{ - 1}}\tan x$

$= {\tan ^{ - 1}}\frac{a}{b} - x$

therefore,$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {{{\tan }^{ - 1}}\frac{a}{b}} \right) - \frac{d}{{dx}}(x)$
$= 0 - 1$
$= - 1$

$\frac{dy}{dx} = -1$