`(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.**