# Another trid identity problem

1. Apr 24, 2009

### DecayProduct

1. The problem statement, all variables and given/known data
(1-cosx)/(1+cosx) = (cscx-cotx)^2

2. Relevant equations
cscx = 1/sinx, cotx = cosx/sinx, sin^2x + cos^2x = 1

3. The attempt at a solution

I have tried many different attempts, but I can't seem to make one side like the other. I took the (cscx-cotx)^2 and expanded it to csc^2x - 2cotxcscx+cot^2x = (1-2cosx+cos^2x)/sin^2x. Alternatively, (1-2cosx+cos^2x)/1-cos^2x

I've done every rearrangement I can think of, but I can't figure how to pull one side out of the other. Any help is much appreciated.

2. Apr 24, 2009

### rootX

Substitute cscx = 1/sinx, cotx = cosx/sinx in the right side and get only sin and cos equation on the right side.

I see you already done that:
(1-2cosx+cos^2x)/1-cos^2x = ((1-cosx)/(sinx))^2 is enough

Left side:
(1-cos.x)/(1+cosx)
.. looks like multiplying both num and den by 1-cosx would help

3. Apr 24, 2009

### Integral

Staff Emeritus
Try multiplying top and bottom of the LHS by (1-cos x).