prove the following

Question

2 cosec 2θ cos2θ = cotθ - tanθ

Answer 1

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!