# One more trig identity

1. May 5, 2009

### chops369

1. The problem statement, all variables and given/known data
I'm finally starting to understand proving trig identities, but I have just one more that I can't seem to figure out.

secx - tanxsinx = cosx

2. Relevant equations
N/A

3. The attempt at a solution
Well first, I multiplied the tanx and sinx and came up with sin2x / cosx
Now I'm stuck. I'm trying to do 1/cosx - sin2x / cosx
Would the GCF be (cosx)(cosx) or just cosx, I'm drawing a blank here. I'm debating whether or not to change the sin2x to 1 - cos2x, but even if I do, I can't figure out how that would simplify to just cosx

2. May 5, 2009

### Cyosis

You're almost there$$\frac{1}{\cos x}-\frac{\sin^2x}{\cos x}=\frac{1}{\cos x}(1-\sin^2x)$$. You should notice something very familiar now.

3. May 5, 2009

### chops369

So would it simplify to cos2x / cosx, which would then simplify to cosx for my answer?

4. May 5, 2009

### Cyosis

Yep it's that easy!