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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.
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