Question

How do you prove `(1+sinx)/(1-sinx)=(secx+tanx)^2`?

Answer 1

see below

Explanation:

`(1+sinx)/(1-sinx)=(secx+tanx)^2`

Right Side `=(secx+tanx)^2`

`=(secx+tanx)(secx+tanx)`

`=sec^2x+2secxtanx+tan^2x`

`=1/cos^2x +2*1/cosx *sinx/cosx +sin^2x/cos^2x`

`=(1+2sinx+sin^2x)/cos^2x`

`=((1+sinx)(1+sinx))/(1-sin^2x)`

`=((1+sinx)(1+sinx))/((1+sinx)(1-sinx))`

`=(1+sinx)/(1-sinx`

`=` Left Side

Answer 2

A simpler derivation is given below :

Explanation:

`{1+sin x}/{1-sin x} = {1+sin x}/{1-sin x} times {1+sin x}/{1+sin x} = {(1+sin x)^2}/{1-sin^2 x} = {(1+sin x)^2}/{cos^2x} = ({1+sin x}/{cos x})^2 = (sec x + tan x)^2`