Prove tan3a= (cos(2a)-cos(4a))/ (sin(4a)- sin(2a))

The solution to the problem is as follows:

cos2A - cos4A = -2 sin(6A/2).sin(-2A/2) = +2 sin(3A).sinA
 
and
 
sin4A - sin2A = 2 cos(6A/2).sin(2A/2) = 2 cos(3A).sinA

Hence RHS = (cos(2A)-cos(4A))/ (sin(4A)- Sin(2A)) = sin 3A / cos 3A = tan 3A = LHS


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Ithink it's the first one ( let me know if it right)

The answer should be c.