[tex]\dfrac{1+\sin 2x}{\cos 2x} = \dfrac{\sin^2 x+\cos^2 x+2\sin x\cos x}{\cos^2 x-\sin^2 x}=\\ \\ = \dfrac{(\sin x+\cos x)^2}{(\cos x + \sin x)(\cos x - \sin x)} = \dfrac{\sin x +\cos x}{\cos x-\sin x}\cdot \dfrac{\dfrac{1}{\cos x}}{\dfrac{1}{\cos x}} = \\ \\\\ = \dfrac{\tan x+1}{1-\tan x} =\dfrac{2+1}{1-2} = \boxed{-3}[/tex]
