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Math Help - Proving

  1. #1
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    Proving

    \displaystyle \begin{align*} \frac{2\tan{x}}{1 - \tan^2{x}} = \frac{2}{\cot{x} - \tan{x}} \end{align*}

    Please prove.
    Last edited by gashbell1; September 19th 2012 at 07:30 AM.
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  2. #2
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    Re: Proving

    Quote Originally Posted by gashbell1 View Post
    Please prove.

    2tan x 2
    -------- = --------
    1- tan2x cot x - tan x

    please prove that L = R


    ??? =? R

    Thanks
    Is this what you're trying to prove?

    \displaystyle \begin{align*} \frac{2\tan{x}}{1 - \tan^2{x}} = \frac{1}{\cot{x} - \tan{x}} \end{align*}
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  3. #3
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    Re: Proving

    Edited. Thanks
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  4. #4
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    Re: Proving

    Quote Originally Posted by gashbell1 View Post
    \displaystyle \begin{align*} \frac{2\tan{x}}{1 - \tan^2{x}} = \frac{2}{\cot{x} - \tan{x}} \end{align*}

    Please prove.
    \displaystyle \begin{align*} RHS &= \frac{2}{\cot{x} - \tan{x}} \\ &= \frac{2}{\frac{1}{\tan{x}} - \tan{x}} \\ &= \frac{2}{\frac{1 - \tan^2{x}}{\tan{x}}} \\ &= \frac{2\tan{x}}{1 - \tan^2{x}} \\ &= LHS \end{align*}
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  5. #5
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    Re: Proving

    But what about left to right?
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  6. #6
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    Re: Proving

    Quote Originally Posted by gashbell1 View Post
    But what about left to right?
    That is redundant. To show two things are equal, you can go either way. It is always best to choose whichever way is easier.
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  7. #7
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    Re: Proving

    Hello, gashbell1!

    \displaystyle\frac{2\tan{x}}{1 - \tan^2{x}} \;=\; \frac{2}{\cot{x} - \tan{x}}

    Left-to-Right:

    We have: . \frac{2\tan x}{1 - \tan^2x}


    Divide numerator and denominator by \tan x

    . . . \dfrac{\dfrac{2\tan x}{\tan x}}{\dfrac{1}{\tan x} - \dfrac{\tan^2x}{\tan x}} \;=\; \dfrac{2}{\cot x - \tan x}
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