Question

Prove : (sec θ - tan θ )^2 = 1 - sin θ /1+sin θ

Answer 1

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