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

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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.**