$\quad \dfrac{1 -2\cos^2x}{\sin^2x\cos^2x}$
$=\dfrac{\sin^2x + \cos^2x - 2\cos^2x}{\sin^2x\cos^2x}$
$=\dfrac{\sin^2x - \cos^2x}{\sin^2x\cos^2x}$
$=\dfrac{\sin^2x}{\sin^2x.\cos^2x} - \dfrac{\cos^2x}{\sin^2x.\cos^2x}$
$=\dfrac{1}{\cos^2x} - \dfrac{1}{\sin^2x}$
$= \tan^2x + 1 - (\cot^2x +1)$
$=\tan^2x - \cot^2x$