Prove that (4tanx(1-tanx)square)/((1+tanx)square)square=sin4x

Answer 1

The way you have written it is incorrect. It should be

4tanx(1-tan^2(x))
-----------------------
(1+tan^2(x))^2

= 4tanx/(1+tanx^2) * (1-tanx^2)/(1+tanx^2)

= [4sinx/cosx * secx^2][(1-tanx^2)/(1+tanx^2)]

= (4sinx cosx)(1-tanx^2)/secx^2
= 2sin2x (cosx^2-sinx^2)
= 2sin2x cos2x
= sin4x