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Another trid identity problem

  1. Apr 24, 2009 #1
    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. jcsd
  3. Apr 24, 2009 #2
    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
     
  4. Apr 24, 2009 #3

    Integral

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