PreCal.Proving Identities
Calculus

Tutor: None Selected  Time limit: 1 Day 
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/(1cosØ)  (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 wellknown identity
Using descriptions from plane geometry (rightangle 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
sinØsinØcosØ/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) = (1cosØ)(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
Secure Information
Content will be erased after question is completed.