Question

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

Answer 1

$(sec x - tan x)^2 \\ \\ = sec^2x - 2 sec x tan x + tan^2 x \\ \\ =(1+tan^2 x) - 2 sec x tan x +tan^2 x \\ \\ =1 - 2 sec x tan x + 2 tan^2 x \\ \\ = 1 - 2tan x(sec x - tan x) \\ \\ =1 - \frac{2 sin x}{cos x} (\frac{1-sin x}{cos x}) \\ \\ = 1 - \frac{2 sin x (1-sin x)}{cos^2 x} \\ \\ =1 - \frac{2 sin x (1-sin x)}{1-sin^2 x} \\ \\ =1 - \frac{2 sin x (1-sin x)}{(1-sin x)(1+sin x)} \\ \\ =1-\frac{2 sin x}{1+sin x}$