(tan2x-tanx)(sin2x-tanx)=tan²x

Giải thích các bước giải:

 $VT=(\tan2x-\tan x)(\sin2x-\tan x)\\
=\left (\dfrac{\sin2x}{\cos2x}-\dfrac{\sin x}{\cos x}  \right )\left (\sin2x-\dfrac{\sin x}{\cos x}  \right )\\
=\dfrac{\sin2x\cos x-\sin x\cos2x}{\cos2x\cos x}.\dfrac{\sin2x\cos x-\sin x}{\cos x}\\
=\dfrac{2\sin x\cos^2x-\sin x(2\cos^2x-1)}{\cos2x\cos x}.\dfrac{2\sin x\cos^2x-\sin x}{\cos x}\\
=\dfrac{2\sin x\cos^2x-2\sin x\cos^2x+\sin x}{\cos2x\cos x}.\dfrac{\sin x(2\cos^2x-1)}{\cos x}\\
=\dfrac{\sin x}{\cos2x\cos x}.\dfrac{\sin x(2\cos^2x-\sin^2x-\cos^2x)}{\cos x}\\
=\dfrac{\sin x}{\cos2x\cos x}.\dfrac{\sin x\cos2x}{\cos x}\\
=\dfrac{\sin^2x}{\cos^2x}\\
=\tan^2x=VP\Rightarrow ĐPCM$