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Math Help - Prove the identity

  1. #1
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    Prove the identity

    Hi All,
    This reads as
    one over sec theta + tan theta = sec theta - tan theta = cos theta over one + sin theta
    1 / secѳ + tanѳ = sec ѳ tanѳ = cosѳ / 1 + sinѳ
    My job is to prove the identities
    Any help would be appreciated
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  2. #2
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    Quote Originally Posted by Joel View Post
    Hi All,
    This reads as
    one over sec theta + tan theta = sec theta - tan theta = cos theta over one + sin theta
    1 / secѳ + tanѳ = sec ѳ tanѳ = cosѳ / 1 + sinѳ
    My job is to prove the identities
    Any help would be appreciated
    Please either use LaTeX or brackets, so that we know what you are talking about...

    Is the first one \frac{1}{\sec{\theta} + \tan{\theta}} or \frac{1}{\sec{\theta}} + \tan{\theta}?
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    Sorry its the first one
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  4. #4
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    Quote Originally Posted by Joel View Post
    Hi All,
    This reads as
    one over sec theta + tan theta = sec theta - tan theta = cos theta over one + sin theta
    1 / secѳ + tanѳ = sec ѳ tanѳ = cosѳ / 1 + sinѳ
    My job is to prove the identities
    Any help would be appreciated

    Second Identity:

    \frac{1}{\sec{\theta} + \tan{\theta}} = \frac{1}{\frac{1}{\cos{\theta}} + \frac{\sin{\theta}}{\cos{\theta}}}

     = \frac{1}{\frac{1 + \sin{\theta}}{\cos{\theta}}}

     = \frac{\cos{\theta}}{1 + \sin{\theta}}


    First identity:

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

     = \frac{\cos{\theta} - \cos{\theta}\sin{\theta}}{\cos^2{\theta}}

     = \frac{1}{\cos{\theta}} - \frac{\sin{\theta}}{\cos{\theta}}

     = \sec{\theta} - \tan{\theta}.
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  5. #5
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    Thanks heaps, your proofs are very neatly set out and clear.

    Thankyou.
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