# Prove that (1 + sinA) / CosA = (sinA- cosA+1) / (sinA+ cosA -1 )any proof with out cross multiplication

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We have to prove that: (1 + sin A)/cos A = (sin A - cos A + 1)/ (sin A + cos A - 1)

To accomplish this we rewrite (1 + sin A)/cos A = (sin A - cos A + 1)/(sin A + cos A - 1) as (1 + sin A)(sin A + cos A - 1) = (sin A - cos A + 1)(cos A) and see if this is true.

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

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

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

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

which is always true

**This proves that (1 + sin A)/cos A = (sin A - cos A + 1)/ (sin A + cos A - 1)**

We'll use properties of proportions to prove the identity.

First, we'll use componendo property:

`a/b = c/d =gt (a+b)/b = (c+d)/d`

`(1+sinA+cosA)/cosA = (sinA-cosA+1+sinA+cosA-1)/(sinA+cosA-1)`

We'll reduce like terms from the numerator from the right side:

`(1 + sinA + cosA)/cosA= (2sinA)/(sinA + cosA - 1)`

Now, we'll cross multiply:

(sinA+cosA + 1)(sinA + cosA - 1) = 2sinAcosA

We notice that the product from the left side returns a difference of two squares, while, to the right side, we've get a double angle identity:

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

We'll expand the binomial:

`sin^2 A + 2sinAcosA + cos^2A - 1 = sin 2A`

But, from Pythagorean identity, we'll get `sin^2A + cos^2A = 1` .

1 + sin2A - 1 = sin2A

We'll eliminate like terms from the left side:

sin2A = sin2A

**We notice that we've get the same double angle identity both sides, therefore, the given expression represents an identity.**