Prove that tan4θ=(4tanθ-4tan^3θ)/(1-6tan^2θ+tan^4θ)

Steps to prove tan4θ=(4tanθ-4tan^3θ)/(1-6tan^2θ+tan^4θ)

L.H.S

=tan4θ

=tan(3θ+θ)

=(tan3θ+tanθ)/(1-tan3θ.tan^θ)

[tan3A=(3tanA-tan^3A)/(1-3tan^2A)]

={(3tanθ-tan^3θ)/(1-3tan^2θ)+tanθ}/{1-(3tanθ-tan^3θ)/(1-3tan^2θ).tan^θ}

={(3tanθ-tan^3θ+tanθ-3tan^3θ)/(1-3tan^2θ)}/{(1-3tan^2θ-3tan^2θ+tan^4θ)/(1-3tan^2θ)}

=(3tanθ-tan^3θ+tanθ-3tan^3θ)/(1-3tan^2θ-3tan^2θ+tan^4θ)

=(4tanθ-4tan^3θ)/(1-6tan^θ+tan^4θ)

=R.H.S

Hence tan4θ=(4tanθ-4tan^3θ)/(1-6tan^2θ+tan^4θ) is proved.

Detail Information:-

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