Chứng minh $\frac{1-2{\sin }^{2}x}{1+\sin 2x}=\frac{1-\tan x}{1+\tan x}$

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

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

Từ (1) và (2) suy ra: $\frac{1-2{\sin }^{2}x}{1+\sin 2x}=\frac{1-\tan x}{1+\tan x}$