+466 I need an answer to this question when you get the chance 
http://web2.0calc.com/questions/trigonometric-identities_8090
+2592 Sorry man. I got it simplified down to
1/(sin(x)*cos(x))=cot(x)+tan(x)
Maybe this?
csc(x)/cos(x) = cot(x) + tan(x) ?
+128578 2 csc 2x = cot x + tan x
2/ [sin (2x) ] = cosx/sinx + sinx/ cosx
Get a common denominator on the right, use an identity on the left
2 / [ sinxcos x + sinxcosx] = [ cos^2x + sin^2x] / [sinx cosx]
2/ [ 2sinxcosx] = 1/ [sinxcosx] the 2's on the LHS "cancel".....and we're left with
1/ [sinxcosx] = 1/ [sinxcosx]
