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Math Help - verifying identities attempt 3...

  1. #1
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    verifying identities attempt 3...

    Here is the problem:

    \frac{tan-cot}{tan+cot}+1 = 2sin^{2}

    I convert it into terms of cos and sin and try to simplify the left side.

    <br />
\frac{sin}{cos}-\frac{cos}{sin}
    _________

    \frac{sin}{cos}+\frac{cos}{sin}

    Then, if I took it a step further and made them get a denominator of cos sin they'd end up on top with

    sin^{2}-cos^{2} and on the bottom sin^{2}+cos^{2}. I could make the bottom (using the Pythagorean identity):

    \frac{1}{cos sin}

    And that's how far I've gotten before everything fails.
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  2. #2
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    Quote Originally Posted by A Beautiful Mind View Post
    Here is the problem:

    \frac{tan-cot}{tan+cot}+1 = 2sin^{2}

    I convert it into terms of cos and sin and try to simplify the left side.

    <br />
\frac{sin}{cos}-\frac{cos}{sin}
    _________

    \frac{sin}{cos}+\frac{cos}{sin}

    Then, if I took it a step further and made them get a denominator of cos sin they'd end up on top with

    sin^{2}-cos^{2} and on the bottom sin^{2}+cos^{2}. I could make the bottom (using the Pythagorean identity):

    \frac{1}{cos sin}

    And that's how far I've gotten before everything fails.

    So far it sounds like you've ended up with

    \frac{\sin^2{x} - \cos^2{x}}{\sin^2{x} + \cos^2{x}} + 1

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

     = \sin^2{x} - 1 + \sin^2{x} + 1

     = 2\sin^2{x}.
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