Chứng minh đẳng thức : $\frac{sin2x-2sinx}{sin2x+2sinx}$ =- $tan^{2}$ $\frac{x}{2}$

Ta có

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

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

$= \dfrac{\cos x - 1}{\cos x + 1}$

Lại có

$\tan^2 \left( \dfrac{x}{2} \right) = \sin^2 \left( \dfrac{x}{2} \right) : \cos^2 \left( \dfrac{x}{2} \right)$

$= \dfrac{1 - \cos x}{2} : \dfrac{1 + \cos x}{2}$

$= \dfrac{1-\cos x}{1 + \cos x}$

Suy ra

$-\tan^2 \left( \dfrac{x}{2} \right) = \dfrac{\cos x - 1}{\cos x + 1} = \dfrac{\sin(2x) - 2\sin x}{\sin(2x) + 2\sin x}$.