Prove

$\frac{1}{\cos ^{2} x(1-\tan x)^{2}}=\frac{\sec ^{2} x}{(1-\tan x)^{2}}$

Let $(1-\tan x)=t$

$\therefore-\sec ^{2} x d x=d t$

$\Rightarrow \int \frac{\sec ^{2} x}{(1-\tan x)^{2}} d x=\int \frac{-d t}{t^{2}}$

$=-\int t^{-2} d t$

$=+\frac{1}{t}+\mathrm{C}$

$=\frac{1}{(1-\tan x)}+\mathrm{C}$