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Math Help - trig identity

  1. #1
    Member smmmc's Avatar
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    trig identity

    Hey
    Prove: cos(x)/ 1-tan(x) + sin(x)/ 1-cot(x) = sin(x) + cos(x)

    i subbed 1-tan(x)= sin(x)/cos(x)
    and cot(x) = cos(x)/sin(x)

    so i divided it and times it all out

    and i get to this stage

    1/ 2cos(x)sin(x)-cos^2(x)-sin^2(x) and didnt knwo what to do

    Thanks
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  2. #2
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    Hi smmmc
    Quote Originally Posted by smmmc View Post
    i subbed 1-tan(x)= sin(x)/cos(x)
    You forgot the term "1"

    1-\tan(x) = 1-\frac{\sin(x)}{\cos(x)}
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  3. #3
    Member smmmc's Avatar
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    yerh i did that.
    Let me just show you my whole working out. sorry its not in latex, dont know how to use it

    So,

    cos(x)/(1- sin(x)/cos(x)) + sin(x)/(1-cos(x)/sin(x))

    = cos(x)/(cos(x)-sin(x)/cos(x)) + sin(x)/(sin(x)-cos(x)/sin(x))

    = cos(x) * cos(x)/(cos(x)-sin(x)) + sin(x) * (sin(x)/(sin(x)-cos(x))

    = cos^2(x) + sin^2(x)/ (cos(x)-sin(x)*(sin(x)/(sin(x)-cos(x))

    which equals

    1/ (2cos(x)sin(x)-cos^2(x)-sin^2(x))

    help please
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  4. #4
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    Hi smmmc
    Quote Originally Posted by smmmc View Post
    cos^2(x) + sin^2(x)/ (cos(x)-sin(x)*(sin(x)/(sin(x)-cos(x))
    What's this ??

    cos(x) * cos(x)/(cos(x)-sin(x)) + sin(x) * (sin(x)/(sin(x)-cos(x))
    It's good till this part. Then, you can manipulate it like this :

    \frac{\cos^{2}(x)}{\cos(x)-sin(x)}+\frac{sin^{2}(x)}{\sin(x) - cos(x)}

    = \frac{\cos^{2}(x)}{\cos(x)-sin(x)} - \frac{sin^{2}(x)}{\cos(x) - sin(x)}

    And a little suggestion from me : How about learning latex ? http://www.mathhelpforum.com/math-he...-tutorial.html
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  5. #5
    Member smmmc's Avatar
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    so it then becomes

    cos^2(x)-sin^2(x)/ cox(x)-sin(x)

    can you cross it out? simplify?

    but that leads to cos(x)-sin(x).. and i dint know you could simplify like that?

    help
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  6. #6
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    Hi smmmc

    \frac{\cos^{2}(x)-sin^{2}(x)}{\cos(x)-sin(x)} doesn't simplify to \cos(x) - \sin(x)

    \cos^{2}(x)-sin^{2}(x) = (\cos x - \sin x)  (\cos x+\sin x)
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