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[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 ;)