$\dfrac{1-2\sin^2x}{1+\sin2x}$
$=\dfrac{(1-\sin^2x-\sin^2x}{\sin^2x+2\sin x\cos x+\cos^2x}$
$=\dfrac{\cos^2x-\sin^2x}{(\sin x+\cos x)^2}$
$=\dfrac{(\cos x-\sin x)(\cos x+\sin x)}{(\cos x+\sin x)^2}$
$=\dfrac{\cos x-\sin x}{\cos x+\sin x}$
$=\dfrac{1-\tan x}{1+\tan x}$