prove that sin^4x-cos^4x/cotx= tanx-2cosxsinx
sin^4x - cos^4x = (sin^2x - cos^2x)(sin^2x+cos^2x) = sin^2x - cos^2x
so, now you have
(sin^2x - cos^2x)/cotx = tanx - 2sinx cosx
sin^2x - cos^2x = 1 - 2cos^2x
-cos2x = -cos2x
I will assume you mean:
(sin^4x-cos^4x)/cotx= tanx-2cosxsinx
LS = (sin^4x-cos^4x)/cotx
= (sin^2 x + cos^2 x)(sin^2 x - cos^2 x)tanx
= (1)(sin^2 x - cos^2 x)tanx
= -cos(2x) tanx
RS = tanx - 2sinxcosx
= sinx/cosx - 2sinxcosx
= (sinx - 2sinx cos^2 x)/cosx
= sinx(1 - 2cos^2 x)/cosx
= tanx(-cos(2x))
= LS
Not claiming this is the easiest way, but it worked out ok