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

can someone please show me the steps to making LEFT SIDE = RIGHT SIDE?

i can't seen to solve this equation for the life of me :(

working with left side notice we have double angles here so we need to use double angle identities:
sin2x = 2sinxcosx
cos2x = cos²x-sin²x

[2(cosxsin2x - sinxcos2x)] / sin2x

[2cosx(2sinxcosx) - 2sinx(cos²x-sin²x)]/2sinxcosx

[4cos²xsinx - 2sinxcos²x+2sin³x]/2sinxcosx

[2sinx(2cos²x - cos²x + sin²x)]/2sinxcosx

(cos²x + sin²x)/cosx

1/cosx

secx