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Homework Help: Tricky Trig Problem

  1. Sep 3, 2010 #1

    Char. Limit

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    1. The problem statement, all variables and given/known data
    All right, so I was trying to help a friend prove a certain (complicated) trig identity for summer homework, but I got stuck myself... hopefully one of you will be able to help.

    The trig identity in question is...

    [tex]\frac{cos(x)}{1-tan(x)} + \frac{sin(x)}{1-cot(x)} = cos(x) + sin(x)[/tex]


    2. Relevant equations
    1+tan^2(x)=sec^2(x)
    1+cot^2(x)=csc^2(x)


    3. The attempt at a solution

    So far I've gotten it to...

    [tex]\frac{cos(x)-sin(x)}{sec^2(x)-2tan(x)} - \frac{cos(x)-sin(x)}{csc^2(x)-2cot(x)} = cos(x)+sin(x)[/tex]

    But although I think that's a really nice form (two very similar terms), I have no idea where to go from there. Could one of you help me out?
     
  2. jcsd
  3. Sep 3, 2010 #2

    Dick

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    I would first express all of the trig functions in terms of sin(x) and cos(x) and then show the two sides are equal. It's pretty straightforward.
     
  4. Sep 3, 2010 #3

    eumyang

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    I would multiply the first fraction by
    [tex]\frac{1 + \tan \,x}{1 + \tan \,x}[/tex]
    and multiply the second fraction by
    [tex]\frac{1 + \cot \,x}{1 + \cot \,x}[/tex].



    69
     
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