Prove that {2sinx} / {sin2xcosx} = sec^2x is a true identity

Please help???

{2sinx} / {sin2xcosx}
= {2sinx} / {2sinxcosxcosx} because of the trig identity sin2x = 2sinxcosx
= 1 / (cosxcosx) cancel the 2sinx (one on top one on bottom
= 1 / cos^2 x self explanatory
= sec^2 x because secant is the reciprocal of cos

To prove identities like this, it's easiest to start with the more complex side and simplify it until you get to the other side. So, I'll start with:
2sinx/sin2xcosx
You should know that sin2x=2sinxcosx.
2sinx/2sinxcosxcosx
2sinx/2sinxcos²x
The 2sinx/2sinx cancels, leaving
1/cos²x
Since 1/cosx=secx, and 1/cos²x=1/cosx*1/cosx,
1/cos²x=secx*secx
Which equals
sec²x. The desired results ahve been achieved.

{2sinx} / {sin2xcosx} =
2sinx/2sinxcosxcosx =
1/cos^2(x) = sec^2(x)

{2sinx} / {sin2xcosx} = {2sinx} / {2sinxcos²x} = 1 / cos²x = sec²x. True.