1. Jun 4, 2008 #1

   thomas49th

   

   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 ;)

    

   thomas49th, Jun 4, 2008
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3. Jun 4, 2008 #2

   Gib Z

   

   Homework Helper

   Multiply every term in the numerator and the denominator by cos^2 x.

    

   Gib Z, Jun 4, 2008
4. Jun 7, 2008 #3

   Dylanette

   

   What do you need to prove about it ? Generally you prove a proposition. Whats your proposition ?

    

   Dylanette, Jun 7, 2008
5. Jun 8, 2008 #4

   Gib Z

   

   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]

    

   Gib Z, Jun 8, 2008

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