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Hard trig proof

  1. Jun 4, 2008 #1
    I have to prove this:

    [tex]\frac{sec^{2}A}{1-tan^{2}A}\equivsec2A[/tex]

    I have got it down to this:

    [tex]\frac{1+\frac{sin^{2}A}{cos^{2}A}}{1-\frac{sin^{2}A}{cos^{2}A}}[/tex]

    and the book says it can be then further equated to

    [tex]\frac{cos^{2}A+sin^{2}A}{cos^{2}A-sin^{2}A}[/tex]

    but i cant see how that is done :(

    Can someone show me

    Thanks ;)
     
  2. jcsd
  3. Jun 4, 2008 #2

    Gib Z

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    Homework Helper

    Multiply every term in the numerator and the denominator by cos^2 x.
     
  4. Jun 7, 2008 #3
    What do you need to prove about it ? Generally you prove a proposition. Whats your proposition ?
     
  5. Jun 8, 2008 #4

    Gib Z

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    Homework Helper

    There was an error in his latex code, if you know how to read latex code click on his latex and you'll see he meant

    [tex]
    \frac{\sec^{2}A}{1-\tan^{2}A} = \sec2A
    [/tex]
     
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