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Thread: Proving

  1. January 19th 2010, 10:25 PM #1
    Punch
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    Proving

    Prove <br /> \frac{1+tan^2A}{1-tan^2A}=sec2A<br />
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  2. January 19th 2010, 11:22 PM #2
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    Quote Originally Posted by Punch View Post
    Prove <br /> \frac{1+tan^2A}{1-tan^2A}=sec2A<br />
    \frac{1 + \tan^2{x}}{1 - \tan^2{x}} = \frac{1 + \frac{\sin^2{x}}{\cos^2{x}}}{1 - \frac{\sin^2{x}}{\cos^2{x}}}

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

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

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

    = \frac{1}{\cos{2x}}

    = \sec{2x}.
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