Ta có
$2VP = cot(2x) - \tan(2x)$
$= \dfrac{\cos(2x)}{\sin(2x)} - \dfrac{\sin(2x)}{\cos(2x)}$
$= \dfrac{\cos^2(2x) - \sin^2(2x)}{\sin(2x) \cos(2x)}$
Vậy
$VP = \dfrac{\cos(4x)}{\sin(4x)}$
$= \dfrac{2\cos^2(4x)}{2\sin(4x) \cos(4x)}$
$= \dfrac{1 + \cos(8x)}{\sin(8x)}$
Vậy đề đúng phải là
$\dfrac{1 + \cos(8x)}{\sin(8x)} = \dfrac{1}{2} \left[ cot(2x) - \tan(2x) \right]$
