Chứng minh rằng tan²x + cot²x = $\frac{6+2\cos 4x}{1-\cos 4x}$ .

tan²x + cot²x = $\frac{{\sin }^{2}x}{{\cos }^{2}x}+\frac{{\cos }^{2}x}{{\sin }^{2}x}=\frac{{\sin }^{4}x+{\cos }^{4}x}{{\sin }^{2}x.{\cos }^{2}x}$

= $\frac{\left({\sin }^{2}x+{\cos }^{2}x\right)-2{\sin }^{2}x.{\cos }^{2}x}{{\sin }^{2}x.{\cos }^{2}x}$

= $\frac{1-2{\sin }^{2}x.{\cos }^{2}x}{{\sin }^{2}x.{\cos }^{2}x}$

= $\frac{1-\frac{1}{2}{\sin }^{2}2x}{\frac{1}{4}{\sin }^{2}2x}$

= $\frac{1-\frac{1}{2}.\frac{1-\cos 4x}{2}}{\frac{1}{4}.\frac{1-\cos 4x}{2}}$

= $\frac{1-\frac{1}{4}.\left(1-\cos 4x\right)}{\frac{1}{8}.\left(1-\cos 4x\right)}$

= $\frac{8-2+2\cos 4x}{1-\cos 4x}$

= $\frac{6+2\cos 4x}{1-\cos 4x}$

Vậy tan²x + cot²x =  $\frac{6+2\cos 4x}{1-\cos 4x}$