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

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.