Chứng minh rằng: $\frac{sin^23x}{sin^2x}$ -$\frac{cos^23x}{cos^2x}$ =8cos2x

$VT=\dfrac{\sin^23x.\cos x-\sin^2x\cos^23x}{\sin^2x\cos^2x}$

$=\dfrac{(\sin3x.\cos x)^2-(\cos3x.\sin x)^2}{\sin^2x\cos^2x}$

$=\dfrac{(\sin3x\cos x-\cos3x\sin x)(\sin3x\cos x+\cos3x\sin x)}{\sin^2x\cos^2x}$

$=\dfrac{\sin2x.\sin4x}{\sin x\cos x\sin x\cos x}$

$=\dfrac{2\sin x\cos x.2\sin2x\cos2x}{(\sin x\cos x)(\sin x\cos x)}$

$=\dfrac{4.2\sin x\cos x.\cos2x}{\sin x\cos x}$

$=8\cos2x$

$=VP$