Question

Can anyone help me to prove this identity？


`(1 + sin \: x)/(1 - sin \: x) = \frac{1}{(sec \: x \: - tan \: x) {}^(2) }`

Answer 1

`(1 + \sin x)/(1 - \sin x) = (1)/((\sec x - \tan x)^2 ) \\\\(1-\sin^2 x)/((1-\sin x)^2)=(1)/(\left((1)/(\cos x) -(\sin x)/(\cos x)\right)^2 )\\\\(\cos^2 x)/((1-\sin x)^2 )=(1)/(\left((1-\sin x)/(\cos x)\right)^2)\\\\(\cos^2 x)/((1-\sin x)^2)=(1)/(((1-\sin x)^2)/(\cos^2 x))\\\\(\cos^2 x)/((1-\sin x)^2)=(\cos^2 x)/((1-\sin x)^2)`