# Hard trig proof

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 cant see how that is done :(

Can someone show me

Thanks ;)

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