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Math Help - Trig Identity Yay!

  1. #1
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    Trig Identity Yay!

    Hey there. I have been trying to prove this trig identity is true for quite some time now with no success.

    Here it is:

    1-cosx/sinx + sinx/1-cosx = 2cscx

    I am confused on how to elminate the fractions any insight would be appreciated thanks.
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  2. #2
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    Quote Originally Posted by Jkeeb View Post
    Hey there. I have been trying to prove this trig identity is true for quite some time now with no success.

    Here it is:

    1-cosx/sinx + sinx/1-cosx = 2cscx

    I am confused on how to elminate the fractions any insight would be appreciated thanks.
    \dfrac{1-\cos{x}}{\sin{x}} + \dfrac{\sin{x}}{1-\cos{x}} =<br />

    common denominator ...

    \dfrac{(1-\cos{x})(1-\cos{x}) + (\sin{x})(\sin{x})}{\sin{x}(1-\cos{x})} =<br />

    \dfrac{1-2\cos{x}+\cos^2{x} + \sin^2{x}}{\sin{x}(1-\cos{x})} =

    \dfrac{2-2\cos{x}}{\sin{x}(1-\cos{x})} =

    \dfrac{2(1-\cos{x})}{\sin{x}(1-\cos{x})} = ... you're there.
    Last edited by skeeter; February 13th 2011 at 06:42 PM. Reason: fixed latex
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  3. #3
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    Hello, Jkeeb!

    Another approach . . .


    \dfrac{1-\cos x}{\sin x} + \dfrac{\sin x}{1-\cos x} \:=\: 2\csc x

    Multiply the second fraction by \frac{1+\cos x}{1 + \cos x}

    . . \displaystyle \frac{1-\cos x}{\sin x} + \frac{\sin x}{1-\cos x}\cdot\frac{1+\cos x}{1 + \cos x} \;=\; \frac{1-\cos x}{\sin x} + \frac{\sin x(1 + \cos x)}{1 - \cos^2\!x}

    . . \displaystyle =\;\frac{1-\cos x}{\sin x} + \frac{\sin x(1 + \cos x)}{\sin^2\!x} \;=\;\frac{1-\cos x}{\sin x} + \frac{1 + \cos x}{\sin x}

    . . \displaystyle =\;\frac{(1 - \cos x) + (1+\cos x)}{\sin x} \;=\;\frac{2}{\sin x} \;=\;2\csc x

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