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sgmartin
Nov 5, 2009, 04:27 PM
sin2x/sinx - cos2x/cosx = secx
ebaines
Nov 5, 2009, 05:17 PM
Remember these important identities:

sin2x = 2sinxcosx
cos2x = cos^2x - sin^x
= 2 cos^x - 1
= 1-2sin^2x.

That second identity for cos2x looks promising, since you will be dividing by cosx. Try it, and see if you make progress.