prove the following
2 cosec 2θ cos2θ = cotθ - tanθ
Identities (I'll use "x" for theta):
sinx/cosx + cosx/sinx = 2cosec2x
cos^2x - sin^2x = cos2x
sin^2x + cos^2x = 1
Working with the left side:
(sinx/cosx + cosx/sinx)(cos^2x - sin^2x) =
(sin^2x/sinxcosx + cos^2x/sinxcosx)(cos^2x - sin^2x) =
[(sin^2x + cos^2x)/sinxcosx](cos^2x - sin^2x) =
(1/sinxcosx)(cos^2x - sin^2x) =
(cos^2x/sinxcosx) - (sin^2x/sinxcosx) =
cosx/sinx - sinx/cosx =
cotx - tanx
If I haven't missed anything, that should be it!