$VT=\dfrac{1+\cos x}{\sin x}.\left[ 1-\dfrac{(1-\cos x)^2}{\sin^2x}\right]$
$=\dfrac{1+\cos x}{\sin x}-\dfrac{1+\cos x}{\sin x}.\dfrac{(1-\cos x)^2}{\sin^2x}$
$=\dfrac{1+\cos x}{\sin x}-\dfrac{(1+\cos x)(1-\cos x)(1-\cos x)}{\sin^3x}$
$=\dfrac{1+\cos x}{\sin x}-\dfrac{(1-\cos^2x)(1-\cos x)}{\sin^2x.\sin x}$
$=\dfrac{1+\cos x}{\sin x}-\dfrac{1-\cos x}{\sin x}$
$=\dfrac{1+\cos x-1+\cos x}{\sin x}$
$=\dfrac{2\cos x}{\sin x}$
$=2\cot x$
$=VP$
