Giải thích các bước giải:
$VT=\dfrac{1-\cos x}{\sin x}.\left ( \dfrac{(1+\cos x)^2}{\sin^2x}-1 \right )\\
=\dfrac{1-\cos x}{\sin x}.\left ( \dfrac{(1+\cos x)^2}{\sin^2x}-\dfrac{\sin^2x}{\sin^2x} \right )\\
=\dfrac{1-\cos x}{\sin x}.\dfrac{\cos^2x+\sin^2x+2\cos x+\cos^2x-\sin^2x}{\sin^2x}\\
=\dfrac{1-\cos x}{\sin x}.\dfrac{2\cos^2x+2\cos x}{\sin^2x}\\
=\dfrac{1-\cos x}{\sin x}.\dfrac{2\cos x(\cos x+1)}{\sin^2x}\\
=\dfrac{2\cos x(1-\cos^2x)}{\sin^3x}\\
=\dfrac{2\cos x\sin^2x}{\sin^3x}\\
=\dfrac{2\cos x}{\sin x}\\
=2\cot x=VP\Rightarrow đpcm$
