Giải thích các bước giải:
$\dfrac{\sin^2x}{\sin x-\cos x}+\dfrac{\sin x+\cos x}{\tan^2x-1}$
$=\dfrac{\sin^2x}{\sin x-\cos x}+\dfrac{\sin x+\cos x}{\dfrac{\sin^2x}{\cos^2x}-1}$
$=\dfrac{\sin^2x}{\sin x-\cos x}+\dfrac{\sin x+\cos x}{\dfrac{\sin^2x-\cos^2x}{\cos^2x}}$
$=\dfrac{\sin^2x}{\sin x-\cos x}+\dfrac{\cos^2x}{\sin x-\cos x}$
$=\dfrac{\sin^2x+\cos^2x}{\sin x-\cos x}$
$=\dfrac{1}{\sin x-\cos x}$