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Math Help - Prove the identity plz

  1. #1
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    Prove the identity plz

    I have been working on this problem for quite some time and have gotten to the point that I am stuck. any way to prove it would be greatly appreciated.

    1
    secx+tanx

    =

    1-sinx
    cosx
    Last edited by nightrider456; January 11th 2010 at 03:32 PM.
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  2. #2
    Super Member bigwave's Avatar
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    Quote Originally Posted by nightrider456 View Post
    I have been working on this problem for quite some time and have gotten to the point that I am stuck. any way to prove it would be greatly appreciated.

    1
    secx=tanx

    =

    1-sinx
    cosx
    first of all

     <br />
\sec{x} = \frac{1}{\cos{x}}<br />

    the way you have this posted is ambiguous (but looks like an easy solution somewhere)
    Last edited by bigwave; January 11th 2010 at 02:37 PM. Reason: wording
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  3. #3
    MHF Contributor ebaines's Avatar
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    solution

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

    <br />
\frac {cos(x) - cos(x)sin(x) } {cos^2(x)} = \frac {1 - sin(x)} {cos(x)}<br />
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  4. #4
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    \frac{1}{\sec{x}+\tan{x}} \cdot \frac{\sec{x}-\tan{x}}{\sec{x}-\tan{x}} =

    \frac{\sec{x} - \tan{x}}{\sec^2{x} - \tan^2{x}} =

    \sec{x} - \tan{x} =

    \frac{1}{\cos{x}} - \frac{\sin{x}}{\cos{x}} = <br />

    \frac{1-\sin{x}}{\cos{x}}
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