- #1
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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 ;)
[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 ;)