# Trying to prove a trig identity

1. May 31, 2009

### james_stewart

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

cos^2x-cotx
--------------- = cot^2x
sin^2x-tanx

2. Relevant equations

3. The attempt at a solution

every solution I get gives me a zero, not cot^2

2. May 31, 2009

### tiny-tim

Welcome to PF!

Hi james_stewart! Welcome to PF!

(try using the X2 tag just above the Reply box )
Either write cot= cos/sin, tan = sin/cos,

or just multiply both sides by sin2x-tanx

3. May 31, 2009

### james_stewart

i did and i'm not getting the proper results.

when i convert cot and tan to cos/sin and sin/cos i get

cos^2-cos^2
-------------
sin^2-sin^2

4. May 31, 2009

### tiny-tim

(please use the X2 tag just above the Reply box)
No, you should get cos2 - cos/sin on the top …

5. May 31, 2009

### james_stewart

i did

and on the bottom i get sin2-sin/cos

6. May 31, 2009

### tiny-tim

ok, now put sin2-sin/cos as one fraction (ie with everything over the same denominator), and the same for cos2-cos/sin

7. May 31, 2009

### james_stewart

That's how i did it. but where do i get cot2 from this?

cos2 - cos/sin
--------------
sin2 - sin/cos

8. Jun 1, 2009

### tiny-tim

erm … put sin2 - sin/cos as one fraction (ie with everything over the same denominator), and the same for cos2 - cos/sin

9. Jun 1, 2009

### rl.bhat

You can write numerator as
[cos2xsinx -cosx]/sinx. Then take cos(x) common.
Repeat the same thing for denominator and simplify.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook