Thread: Prove that secx=[2(cosxsin2x-sinxcos2x)]/sin2x

Prove that secx=[2(cosxsin2x-sinxcos2x)]/sin2x

explain it to me fully please details please~!

While you are proving this, where are you getting lost?

Which step are you struggling with?

If you are struggling to begin, I normally suggest using your double angle formulae so that all of the trig functions have an argument of just x instead of 2x.

I have up to:
[2sin(2x-x)]/sin2x
[2sin(2x-x)]/2sinxcosx

Questions:
how do I cancel 2sin(2x-x) so that it equals 1/cosx?

does 2sin(2x-x) really equal 2sin(x) or any number for x for that matter?
so can I cancel out the 2sin(2x-x)/2sinx?

Originally Posted by skeske1234

I have up to:
[2sin(2x-x)]/sin2x
[2sin(2x-x)]/2sinxcosx

Questions:
how do I cancel 2sin(2x-x) so that it equals 1/cosx?

does 2sin(2x-x) really equal 2sin(x) or any number for x?
so can I cancel out the 2sin(2x-x)/2sinx?

yes ...