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