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Hard trig proof

  • Thread starter thomas49th
  • Start date
  • #1
655
0
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 ;)
 

Answers and Replies

  • #2
Gib Z
Homework Helper
3,346
4
Multiply every term in the numerator and the denominator by cos^2 x.
 
  • #3
5
0
What do you need to prove about it ? Generally you prove a proposition. Whats your proposition ?
 
  • #4
Gib Z
Homework Helper
3,346
4
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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