Giải thích các bước giải:
$VT=\dfrac{\sin^2x}{\sin x-\cos x}-\dfrac{\sin x+\cos x}{\tan^2x-1}\\
=\dfrac{\sin^2x}{\sin x-\cos x}-\dfrac{\sin x+\cos x}{\dfrac{1}{\cos^2x}-1-1}\\
=\dfrac{\sin^2x}{\sin x-\cos x}-\dfrac{\cos^2x(\sin x+\cos x)}{1-2\cos^2x}\\
=\dfrac{\sin^2x}{\sin x-\cos x}-\dfrac{\cos^2x(\sin x+\cos x)}{\sin^2x+\cos^2x-2\cos^2x}\\
=\dfrac{\sin^2x}{\sin x-\cos x}-\dfrac{\cos^2x(\sin x+\cos x)}{\sin^2x-\cos^2x}\\
=\dfrac{\sin^2x(\sin x+\cos x)}{\sin^2x-\cos^2x}-\dfrac{\cos^2x(\sin x+\cos x)}{\sin^2x-\cos^2x}\\
=\dfrac{\sin^2x(\sin x+\cos x)-\cos^2x(\sin x+\cos x)}{\sin^2x-\cos^2x}\\
=\dfrac{(\sin x+\cos x)(\sin^2x-\cos^2x)}{\sin^2x-\cos^2x}\\
=\sin x+\cos x=VP\Rightarrow ĐPCM$