Ta có:
$VT=\dfrac{1+\cos x}{\sin x}-\dfrac{\sin x}{1+\cos x}$
$=\dfrac{(1+\cos x)^2}{\sin x(1+\cos x)}-\dfrac{\sin^2x}{\sin x(1+\cos x)}$
$=\dfrac{\cos^2x+2\cos x+1-\sin^2x}{\sin x(1+\cos x)}$
$=\dfrac{\cos^2 x+2\cos x+\cos^2 x}{\sin x(1+\cos x)}$
$=\dfrac{2\cos^2x+2\cos x}{\sin x(1+\cos x)}$
$=\dfrac{2\cos x(\cos x+1)}{\sin x(\cos x+1)}$
$=\dfrac{2\cos x}{\sin x}$
$=2\cot x=VT$
Vậy đẳng thức được chứng minh.