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Pretty difficult trig proof (identity)

  1. Aug 25, 2011 #1
    1. The problem statement, all variables and given/known data
    [itex]\frac{sin\theta}{1-cos\theta} - \frac{cot\theta}{1+cos\theta} = \frac{1-cos^{3}\theta}{sin^{3}\theta}[/itex]


    2. Relevant equations
    Trig identities..


    3. The attempt at a solution
    Basically I got to:
    [itex]\frac{sin\theta+(cos^{2}\theta)(sin\theta)}{sin^{2}\theta}[/itex]
    1. The problem statement, all variables and given/known data

    Is that right up to there, I think not because I cant get passed this lol.


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 25, 2011 #2

    SammyS

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    That doesn't look right.

    Show how you got your result so we can help you.
     
  4. Aug 25, 2011 #3

    PeterO

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    You did notice the cot function did you - or did you misread it as cos?
     
  5. Aug 25, 2011 #4

    PeterO

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    I don't think you were supposed to solve the question for the Original Poster!!!!!
     
    Last edited by a moderator: Aug 25, 2011
  6. Aug 25, 2011 #5
    >_> Oh. Well, I hope my explanation blurb thing helps so that I'm not just blatantly giving the solution without providing any real understanding.
     
  7. Aug 25, 2011 #6

    Mark44

    Staff: Mentor

    PeterO is correct. The Physics Forums rules do not permit a member to post the solution to another member's problem.
     
  8. Aug 26, 2011 #7
    OK, I attached the rest of my work..
     

    Attached Files:

  9. Aug 26, 2011 #8

    PeterO

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    In your 5th line, when you took out a factor of sin(theta) in the numerator, it was not a common factor, as it was in the denominator of a couple of the terms.
     
  10. Aug 26, 2011 #9
    I see now thanks, I got it. Stupid mistakes.
     
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