Chứng minh rằng $\frac{cos^{2}x-sin^{2}x}{cot^{2}x-tan^{2}x}$ = $\frac{1}{8}$(1-cos4x)

Ta có

$VT = \dfrac{\cos^2x – \sin^2x}{cot^2x – \tan^2x}$

$= \dfrac{\sin^2 x \cos^2 x(\cos^2x – \sin^2x)}{\cos^4x – \sin^4x}$

$= \dfrac{\sin^2x \cos^2x (\cos^2x – \sin^2x)}{(\cos^2x – \sin^2x)(\cos^2x + sin^2x)}$

$= \sin^2x \cos^2x$

$= \dfrac{1}{4} \sin^2(2x)$

$= \dfrac{1}{8} [1 – \cos(4x)] = VP$