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Math Help - Need help with proving an identity!

  1. #1
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    Need help with proving an identity!

    I need to verify this identity:

    (tan x/(1 + cos x)) + (sin x/(1 - cos x)) = cot x + (sec s)(csc x)

    I've tried it twice now, and on the left side I keep getting:
    sin^2 x /(cot x + (sec x)(csc x)

    What I'm doing is multiplying the denominator and the numerator of the first fraction by (1 - cos x) to get sin x^2 on the bottom, and doing the same with the second fraction by multiplying by (1 + cos x). Simplifying that give me the right half of the identity under sin ^2 X

    It's supposed to be an identity but I guess I'm screwing it up... can someone try it out and post their work so I can see what I'm doing wrong?

    I'd very much appreciate it.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

    The idea is the right one :
    \frac{\tan x}{1+\cos x}+\frac{\sin x}{1-\cos x}=\frac{\frac{\sin x}{\cos x}(1-\cos x)+(1+\cos x)\sin x}{\sin^2x}=\frac{\frac{1-\cos x}{\cos x}+1+\cos x}{\sin x}=\ldots
    Last edited by flyingsquirrel; April 24th 2008 at 01:46 PM. Reason: "good idea" -> "right idea" :o
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  3. #3
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    Thanks... but I dont quite understand what you are doing between the second and third step, and I don't know what to do after that....
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  4. #4
    Moo
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    Hello,

    Divided by sin(x)
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by flyingsquirrel View Post
    Hi

    The idea is the good one :
    \frac{\tan x}{1+\cos x}+\frac{\sin x}{1-\cos x}=\frac{\frac{\sin x}{\cos x}(1-\cos x)+(1+\cos x)(\sin x)}{\sin^2x}=\frac{\frac{1-\cos x}{\cos x}+1+\cos x}{\sin x}=\ldots
    THe good idea?

    Who would have guessed you were French? Bon idee!
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  6. #6
    Moo
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    Quote Originally Posted by Mathstud28 View Post
    THe good idea?

    Who would have guessed you were French? Bon idee!
    Are you saying that French people can't guess correctly ?
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Are you saying that French people can't guess correctly ?
    Mais non!
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  8. #8
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Kiba View Post
    Thanks... but I dont quite understand what you are doing between the second and third step, and I don't know what to do after that....
    \frac{1-\cos x}{\cos x}=\frac{1}{\cos x}-1 then simplify in the last expression of my first post and you got the result.
    Quote Originally Posted by Mathstud28 View Post
    THe good idea?
    Is there another one ?
    Bonne idee!
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by flyingsquirrel View Post
    \frac{1-\cos x}{\cos x}=\frac{1}{\cos x}-1 then simplify in the last expression of my first post and you got the result.

    Is there another one ?
    Crap...and \text{bonNE idee}\Rightarrow{right} \text{idea}
    Your sentence would be "This idea is the correct/right one"
    Dont worry...today I said ....C'est possible ce(instead of que) je n'ais pas raison"
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  10. #10
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    I still don't get it... and I don't know how to simplify that anymore...
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  11. #11
    o_O
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    Quote Originally Posted by Mathstud28 View Post
    Crap...and \text{bonNE idee}\Rightarrow{right} \text{idea}
    Your sentence would be "This idea is the correct/right one"
    Dont worry...today I said ....C'est possible ce(instead of que) je n'aie pas raison"
    As for the problem:
    \frac{\frac{\sin x}{\cos x}(1-\cos x)+(1+\cos x)\sin x}{\sin^2x}

    Factor out sin x from the numerator and cancel with the denominator:
    = \frac{\frac{1}{\cos x} - 1 + 1 + \cos x}{\sin x}

    Simplify:
    = \frac{\sec x + \cos x}{\sin x} = \frac{\sec x}{\sin x} + \frac{\cos x}{\sin x}

    I'm sure you can see what happen sfrom here.
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by o_O View Post
    As for the problem:
    \frac{\frac{\sin x}{\cos x}(1-\cos x)+(1+\cos x)\sin x}{\sin^2x}

    Factor out sin x from the numerator and cancel with the denominator:
    = \frac{\frac{1}{\cos x} - 1 + 1 + \cos x}{\sin x}

    Simplify:
    = \frac{\sec x + \cos x}{\sin x} = \frac{\sec x}{\sin x} + \frac{\cos x}{\sin x}

    I'm sure you can see what happen sfrom here.
    Mon dieu...quest-ce que tu fait? Arrette! Je sais. Je ne suis pas bon a Francais...fiche-moi!
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  13. #13
    o_O
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    Quote Originally Posted by Mathstud28 View Post
    Mon dieu...qu'est-ce que tu fais? Arrŕte! Je sais. Je ne parle pas bien la franšais...fiche-moi le camp (?)!


    Moo can fix anything else
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  14. #14
    Moo
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    He has quite a rude language in French :/ Not only here I mean..
    "le franšais"
    or "fiche-moi la paix"

    It's perfect
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