Giải thích các bước giải:
Ta có:
$\dfrac{2\cos2x - \sin4x}{2\cos2x + \sin4x}$
$=\dfrac{2\cos2x - 2\sin2x\cos2x}{2\cos2x + 2\sin2x\cos2x}$
$=\dfrac{1 -\sin2x}{1 + \sin2x}$
$=\dfrac{\sin^2x+\cos^2x -2\sin x\cos x}{\sin^2x+\cos^2x +2\sin x\cos x}$
$=\dfrac{\left(\sin x-\cos x\right)^2}{\left(\sin x+\cos x \right)^2}$
$=\dfrac{\left(\dfrac{\sqrt{2}}{2}\sin x-\dfrac{\sqrt{2}}{2}\cos x\right)^2}{\left(\dfrac{\sqrt{2}}{2}\sin x+\dfrac{\sqrt{2}}{2}\cos x \right)^2}$
$=\dfrac{\left(\cos\left(\dfrac{\pi}{4}\right)\sin x-\sin\left(\dfrac{\pi}{4}\right)\cos x\right)^2}{\left(\sin\left(\dfrac{\pi}{4}\right)\sin x+\cos x\cos\left(\dfrac{\pi}{4}\right) \right)^2}$
$=\dfrac{\left(\sin x\cos\left(\dfrac{\pi}{4}\right)-\cos x\sin\left(\dfrac{\pi}{4}\right)\right)^2}{\left(\cos\left(x\right)\cos\left(\dfrac{\pi}{4}\right)+\sin x\sin\left(\dfrac{\pi}{4}\right) \right)^2}$
$=\dfrac{\sin^2\left(x-\dfrac{\pi}{4}\right)}{\cos^2\left(x-\dfrac{\pi}{4}\right)}$
$=\tan^2\left(x-\dfrac{\pi}{4}\right)$
