[tex]\dfrac{1 + \sin x}{1 - \sin x} = \dfrac{1}{(\sec x - \tan x)^2 } \\\\\dfrac{1-\sin^2 x}{(1-\sin x)^2}=\dfrac{1}{\left(\dfrac{1}{\cos x} -\dfrac{\sin x}{\cos x}\right)^2 }\\\\\dfrac{\cos^2 x}{(1-\sin x)^2 }=\dfrac{1}{\left(\dfrac{1-\sin x}{\cos x}\right)^2}\\\\\dfrac{\cos^2 x}{(1-\sin x)^2}=\dfrac{1}{\dfrac{(1-\sin x)^2}{\cos^2 x}}\\\\\dfrac{\cos^2 x}{(1-\sin x)^2}=\dfrac{\cos^2 x}{(1-\sin x)^2}[/tex]
