e) Ta có
$VT = \dfrac{\sin x + \cos x - 1}{1 - \cos x}$
$= \dfrac{[\sin x + (\cos x - 1)][\sin x - (\cos x -1)]}{(1 - \cos x)[\sin x - (\cos x - 1)]}$
$= \dfrac{\sin^2x - (\cos x - 1)^2}{(1 - \cos x)(\sin x - \cos x + 1)}$
$= \dfrac{\sin^2x - \cos^2x + 2 \cos x - 1}{(1 - \cos x)(\sin x - \cos x + 1)}$
$= \dfrac{1 - \cos^2x - \cos^2x + 2\cos x - 1}{(1 - \cos x)(\sin x - \cos x + 1)}$
$= \dfrac{2 \cos x - 2\cos^2x}{(1 - \cos x)(\sin x - \cos x + 1)}$
$= \dfrac{2\cos x(1 - \cos x)}{(1 - \cos x)(\sin x - \cos x + 1)}$
$= \dfrac{2\cos x}{\sin x - \cos x + 1} = VP$
