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