## Trig Identity

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.
j3nn4k
Posts: 2
Joined: Fri Mar 21, 2014 2:07 am
Contact:

### Trig Identity

So I'm working on my homework and cannot figure out how to verify this identity.

The instructions on the worksheet are simply to verify that the identity is true.

The equation is;

(sec(A)-tan(A))^2 = (1-sin(A))/(1+sin(A))

I can work with either the left or the right side to verify it, but neither seems to really be helping me out with this problem.
I've worked though several different starting points and keep hitting dead ends.

If someone could help me or at least help me get started in the right direction that would be wonderful.

buddy
Posts: 197
Joined: Sun Feb 22, 2009 10:05 pm
Contact:

### Re: Trig Identity

I've worked though several different starting points and keep hitting dead ends.
Lots of times that means that you have to work both sides down to some place where they're equal & then write it up down one side and then up the other.
(sec(A)-tan(A))^2 = (1-sin(A))/(1+sin(A))
Make the left all sines and cosines: (1/cos(A) - sin(A)/cos(A))^2 = ((1 - sin(A))/cos(A))^2

Make the right so the denom is in cosine: [(1 - sin(A))/(1 - sin(A))][(1 + sin(A))/(1 - sin(A))] = ((1 - sin(A))^2) / (1 - sin^2(A)) so the denom turns into cos^2(A)

So the left equals to the right. Do the proof by starting on one side, going down to where they're equal, and then going up the other side. (This is how they come up with those proofs in the book that make you think "where did THAT come from???")

Hope that helps.

j3nn4k
Posts: 2
Joined: Fri Mar 21, 2014 2:07 am
Contact:

### Re: Trig Identity

Okay, I think I can figure it out now. Thanks so much for the help!