#1-cos2x =tanxsin2x#
I'll prove using the right hand side of the equation.
From the double angle identities, #sin2x=2sinxcosx#:
#quadquadquadquadquadquadquadquadquad=sinx/cosx * 2sinxcosx#
Combine by multiplying:
#quadquadquadquadquadquadquadquadquad=(2sin^2xcancel(cosx))/cancel(cosx)#
#quadquadquadquadquadquadquadquadquad=2sin^2x#
From the Pythagorean Identities, #sin^2x = 1-cos^2x#:
#quadquadquadquadquadquadquadquadquad=2(1-cos^2x)#
Simplify:
#quadquadquadquadquadquadquadquadquad=2-2cos^2x#
Factor out a #-1#:
#quadquadquadquadquadquadquadquadquad=-1(-2+2cos^2x)#
Since we need #2cos^2x-1# to get #cos2x#, let's rewrite it so that we can get that:
#quadquadquadquadquadquadquadquadquad=-1(-1+2cos^2x-1)#
From the double angle identities, #cos^2x-1 = cos2x#:
#quadquadquadquadquadquadquadquadquad=-1(-1+cos2x)#
Finally:
#1-cos2x=1-cos2x#
We have proved that #1-cos2x =tanxsin2x#.
Hope this helps!
