(1+2sin(x)cos(x))/(cos^2(x)-sin^2(x))

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

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

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

Chia tử, mẫu cho $\cos x$ ta được:

$\dfrac{1+\tan x}{1-\tan x}$