One more trig identity

[chops369]

Homework Statement

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

Homework Equations

N/A

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

 

[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.

 

[chops369]

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

 

[Cyosis]

Yep it's that easy!