Proving \frac{sec^{2}A}{1-tan^{2}A} Equals sec2A

[thomas49th]

I have to prove this:

$$\frac{sec^{2}A}{1-tan^{2}A}\equivsec2A$$

I have got it down to this:

$$\frac{1+\frac{sin^{2}A}{cos^{2}A}}{1-\frac{sin^{2}A}{cos^{2}A}}$$

and the book says it can be then further equated to

$$\frac{cos^{2}A+sin^{2}A}{cos^{2}A-sin^{2}A}$$

but i can't see how that is done :(

Can someone show me

Thanks ;)

 

[Gib Z]

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

 

[Dylanette]

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

 

[Gib Z]

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

$$\frac{\sec^{2}A}{1-\tan^{2}A} = \sec2A$$