# Proving Trigonometric Identites

1. Jan 13, 2009

### meeklobraca

1. The problem statement, all variables and given/known data

tan x + 1/tan x - 1 = sec x + csc x/sec x - csc x

2. Relevant equations

3. The attempt at a solution

Ive tried working with the right side and have gotten as far as sin x/cosxsinx - cos x/cosxsinx

Im not sure if im just out to lunch, or on the right track. Any help would be awesome!

Thanks!

2. Jan 13, 2009

### gabbagabbahey

Your expression is ambiguous the way you have written it. (Use either brackets or LaTeX)

Do you mean $$\frac{\tan(x)+1}{\tan(x)-1}=\frac{\sec(x)+\csc(x)}{\sec(x)-\csc(x)}$$ ???

If so, try working with the LHS: Express everything in terms of sines and cosines and then divide both the numerator and denominator by $\sin(x)$

3. Jan 14, 2009

### meeklobraca

im not getting it. should i turn the 1 into sines or cosines? Or even how do I know to do that?

4. Jan 14, 2009

### NoMoreExams

No you should turn the tan into sin/cos and the find common denominators, cancel, etc. etc.

5. Jan 14, 2009

### meeklobraca

Oh I did that too.

I had sin/cos + cos/cos all divided by sin/cos - cos/cos

6. Jan 14, 2009

### NoMoreExams

Well a/c + b/c = (a+b)/c and also (a/b)/(c/d) = (ad)/(bc) I heard maybe you should use that.

7. Jan 14, 2009

### meeklobraca

Well I heard that gives me sin x cos x + cos^2 x / sin x cos x - cos ^2 x

8. Jan 14, 2009

### NoMoreExams

Umm why haven't you cancelled the cos

9. Jan 14, 2009

### meeklobraca

Your right, now I have sin x + cos x / sin x - cos x ?

10. Jan 14, 2009

### NoMoreExams

Now compare that to what you are trying to get to...

11. Jan 14, 2009

### meeklobraca

I got it. I went with the RHS side though, and converted it to the LHS. Thanks for your help!