Prove that (1 + sinA) / CosA = (1 + sinA+ cosA) / (1 + cosA - sinA)

We'll cross multiply and we'll get the equivalent
expression:

cos A*(1 + sinA+ cosA) = (1 + sin A)*(1 + cosA
- sinA)

We'll remove the
brackets:

cos A + cos A*sin A + (cos A)^2 = 1 + cos A - sin
A + sin A + sin A*cos A - (sin A)^2

We'll eliminate like
terms from the right side:

cos A + cos A*sin A + (cos A)^2
= 1 + cos A + sin A*cos A - (sin A)^2

We'll eliminate the
product cos A*sin A both sides:

cos A  + (cos A)^2 = 1 +
cos A  - (sin A)^2

We'll subtract cos A  - (sin A)^2 both
sides:

cos A + (cos A)^2 - cos A  + (sin A)^2 =
1

We'll eliminate cos A:

(cos
A)^2 + (sin A)^2 = 1

But from Pythagorean identity, we'll
get:

(cos A)^2 + (sin A)^2 =
1

Therefore, the given identity (1 + sinA) /
CosA = (1 + sinA+ cosA) / (1 + cosA - sinA) is
verified.