Asked by Ivan h
Prove that (4tanx(1-tanx)square)/((1+tanx)square)square=sin4x
Answers
Answered by
Steve
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
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
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