Question

PLEASE SOLVE THIS TRIGONOMETRY EQUATION IN BRIEF (1-sin^2 theta)/(1-sin theta)^2=1+sin theta/1-sin theta. HOW???????????????????????????

Answer 1

1- sin²a/ (1- sina)²

ie.,(1²-sin²a)/((1-sina)² (: a instead of theta)

(:a²-b²=(a+b)(a-b))

(1+sina)(1-sina)/(1-sina)²

(1+sina)/(1-sina)

Answer 2

✪ᴀɴsᴡᴇʀ✪

$\to \frac{1-sin^2(\theta)} {[1-sin(\theta)]^2}$

$\to \frac{1^2-[sin(\theta)]^2} {[1-sin(\theta)]^2}$

Using $(a^2-b^2) = (a-b)(a+b)$,

$\to \frac{[1-sin(\theta)][1+sin(\theta)]} {[1-sin(\theta)]^2}$

$\to \frac{[1-sin(\theta)][1+sin(\theta)]} {[1-sin(\theta)][1-sin(\theta)]}$

$\to \frac{1+sin(\theta)} {1-sin(\theta)}$

Hence Proved