Prove the identity. `(1-cosA+sinA)/(1-cosA)=(1+sinA+cosA)/(sinA)`

Expert Answers

An illustration of the letter 'A' in a speech bubbles

`(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.

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial Team