Question

How do you prove `Sec(2x)=sec^2x/(2-sec^2x)`?

Answer 1

Proof below

Explanation:

Double angle formula for `cos`: `cos(2A)=cos^A-sin^a or = 2cos^2A - 1 or = 1 - 2sin^2A`
Applying this:
`sec2x=1/cos(2x)`
`=1/(2cos^2x-1)`, then divide top and bottom by `cos^2x`,
`=(sec^2x)/(2-sec^2x)`