Giải thích các bước giải:
Ta có:
$\dfrac{\sin x+\cos x-1}{1-\cos x}=\dfrac{2\cos x}{\sin x-\cos x+1}$
$\leftrightarrow(\sin x+\cos x-1)(\sin x-\cos x+1)=2\cos x(1-\cos x)$
$\leftrightarrow(\sin x+(\cos x-1))(\sin x-(\cos x-1))=2\cos x-2\cos^2x$
$\leftrightarrow\sin^2x-(\cos x-1)^2=2\cos x-2\cos^2x$
$\leftrightarrow\sin^2x-(\cos^2x-2\cos x+1)=2\cos x-2\cos^2x$
$\leftrightarrow\sin^2x-\cos^2x+2\cos x-1=2\cos x-2\cos^2x$
$\leftrightarrow\sin^2x+\cos^2x=1$ luôn đúng
$\to đpcm$
