Prove the identity. `(1-cosA+sinA)/(1-cosA)=(1+sinA+cosA)/(sinA)`
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`(1-cosA+sinA)/(1-cosA)=(1+sinA+cosA)/(sinA)`
To prove, let's try to simplify left side. To do so, multiply its numerator and denominator by the conjugate of 1 - cosA.
`(1-cosA+sinA)/(1-cosA)*(1+cosA)/(1+cosA)=(1+sinA+cosA)/(sinA)`
`(1-cosA+sinA -cosA + cos^2A + sinAcosA)/(1+cosA-cosA - cos^2A)=(1+sinA+cosA)/(sinA)`
`(1 - cos^2A + sinA+ sinAcosA)/(1-cos^2A)=(1+sinA+cosA)/(sinA)`
Then, apply the Pythagorean identity. So `1 - cos^2A = sin^2A` .
`(sin^2A + sinA + sinAcosA)/(sin^2A)=(1+sinA+cosA)/(sinA)`
Factor out the GCF in the numerator.
`(sinA (sinA + 1 +cosA))/(sin^2A)=(1+sinA+cosA)/(sinA)`
Cancel the common factor between the numerator and denominator.
`(sinA + 1 + cosA)/(sinA)=(1+sinA+cosA)/(sinA)`
Then re-writ the left side as:
`(1+ sinA+cosA)/(sinA)=(1+sinA+cosA)/(sinA) (True)`
The simplified form of the left side is the same with the right side. Hence, this proves that the given equation is an identity.
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