# Pre-Cal.-Proving Identities

*label*Calculus

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How can I prove this identity? I know the answer is "true," but I can't figure out exactly how to arrive at the answer.

sinØ/(1-cosØ) - (1+cosØ)/sinØ = 0

sinØ / (1 + cosØ) = (1 - cosØ) / sinØ

I'll use a simpler, step by step approach, the same as when on solves an algebraic equation:

multiply both sides by sinØ (this gets rid of the denominator on the right):

sin^2Ø / (1 + cosØ) = (1 - cosØ)

sin^2Ø simply means "sine squared of phi"

it is the same as writing (sinØ)^2

next step, multiply both sides by 1 + cosØ

this gets rid of the denominator on the left

sin^2Ø = (1 + cosØ)(1 - cosØ)

sin^2Ø = 1 - cos^2Ø

sin^2Ø + cos^2Ø = 1

which is a well-known identity

Using descriptions from plane geometry (right-angle triangle)

(opp/hyp)^2 + (adj/hyp)^2 = 1

(opposite^2 + adjacent^2)/hypotenuse^2 = 1

Using Pythagorean theorem:

hypotenuse^2 = opp.^2 + adj.^2

we replace the top part

hypotenuse^2 / hypotenuse^2 = 1

which is true.

You have

sinØ-sinØcosØ/sinØsinØcosØ = cosØ/sinØsinØcosØ.

I do not see your reasoning.

At first, I thought you were going for a common denominator, but that would have been (for the denominator)

(1 + cosØ)sinØ = sinØ + sinØcosØ

(did you lose the + along the way?)

Next, you would multiply each side by the missing part of the common denominator (which, in this case, is the denominator from the other side of the = sign)

sinØsinØ / (sinØ + sinØcosØ) = (1-cosØ)(1+cosØ) / (sinØ + sinØcosØ)

When you reach this stage, where each side (the whole side) is over the same common denominator, then you can ignore the denominator.

simple example: let's say k is some fixed number and both sides have it as the denominator

a/k = b/k

then, obviously, the only way this can be true is if a=b (you can now ignore the k)

This idea of dropping the common denominator leaves you with

(sinØ)^2 = 1 - (cosØ)^2

which becomes

sin^2Ø + cos^2Ø = 1

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