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Math Help - Help with proving identities

  1. #1
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    Help with proving identities

    I've been studying for an upcoming math test and have come across two problems that I can't figure out how to solve, help is appreciated.

    prove:

    \frac{1 + \ tan(x)}{\sin x + \cos x}=\frac{1}{\cos(x)}

    and question 2:

    \frac{\ tan^2(x)}{1 - \cos x}=\frac{1}{\cos(x)} + \frac{1}{\cos^2(x)}


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  2. #2
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    Quote Originally Posted by Croton View Post
    I've been studying for an upcoming math test and have come across two problems that I can't figure out how to solve, help is appreciated.

    prove:

    \frac{1 + \ tan(x)}{\sin x + \cos x}=\frac{1}{\cos(x)}

    and question 2:

    \frac{\ tan^2(x)}{1 - \cos x}=\frac{1}{\cos(x)} + \frac{1}{\cos^2(x)}


    1. 1+tan(x) = \frac{cos(x)+sin(x)}{cos(x)}<br />

    Sub that expression into the LHS and do some cancelling



    2.

    <br />
\frac{sin^2(x)}{cos^2(x)(1-cos(x))}

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

    \frac{1-cos^2(x)}{cos^2(x)(1-cos(x)}

    Use the difference of two squares on the numerator and cancel
    Last edited by e^(i*pi); February 10th 2010 at 12:53 PM. Reason: adding white space for clarity
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  3. #3
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    For the first problem, write:

    1 + \tan x = \frac{\cos x}{\cos x} + \frac{\sin x}{\cos x} = \frac{\sin x + \cos x}{\cos x}

    For the second problem, write:

    \frac{\tan ^2 x}{1 - \cos x} = \frac{\sin^2 x}{(1 - \cos x)(\cos^2 x)}

    = \frac{1 - \cos^2 x}{(1 - \cos x)(\cos^2 x)} = \frac{(1 - \cos x)(1 + \cos x)}{(1 - \cos x)(\cos^2 x)} = \frac{1 + \cos x}{\cos^2 x}
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  4. #4
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    Ah so that how you do it, Thanks!
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