Đáp án + Giải thích các bước giải:

$\cot^2 x + \tan^2 x = \dfrac{6 + 2 \cos 4x}{1 - \cos 4x}$ 

$\Leftrightarrow \dfrac{\cos^2 x}{\sin^2 x} + \dfrac{\sin^2 x}{\cos^2 x} = \dfrac{6 + 2(1 - 2\sin^2 2x)}{1 - (1 - 2\sin^2 2x)}$

$\Leftrightarrow \dfrac{\cos^4 x + \sin^4 x}{\sin^2 x \cos^2 x} = \dfrac{6 + 2 - 4\sin^2 2x}{2\sin^2 2x}$

$\Leftrightarrow \dfrac{(\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x}{\sin^2 x \cos^2 x} = \dfrac{8 - 4\sin^2 2x}{2\sin^2 2x}$

$\Leftrightarrow \dfrac{1 - 2\sin^2 x \cos^2 x}{\sin^2 x \cos^2 x} = \dfrac{8 - 4[(2\sin x \cos x)^2]}{2(2\sin x \cos x)^2}$

$\Leftrightarrow \dfrac{1 - 2\sin^2 x \cos^2 x}{\sin^2 x \cos^2 x} = \dfrac{8 - 16\sin^2 x \cos^2 x}{8\sin^2 x\cos^2 x}$

$\Leftrightarrow \dfrac{1 - 2\sin^2 x \cos^2 x}{\sin^2 x \cos^2 x} = \dfrac{1 - 2\sin^2 x \cos^2 x}{\sin^2 x \cos^2 x}(\text{luôn đúng})$

$\Rightarrow \text{ĐPCM}$