tanA+tan(60+A)+tan(60-A)=3tan3A

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your question is wrong , the question must be like

tana+tan(60+a)-tan(60-a)=3tan3a

Let's use the sum-of-angles formula for tangent,
tan(x+y) = (tanx + tany)/(1 - tanx tany), along with
tan60º = √3

tan(60º ± A) = (√3 ± tanA) / (1 ∓ (√3) tanA)
= (1 ± (√3) tanA) (√3 ± tanA) / (1 - 3 tan²A)
= (3 ± 4 tanA + (√3) tan²A) / (1 - 3 tan²A)

tanA + tan(60º + A) - tan(60º - A)
= tanA + 8 tanA / (1 - 3 tan²A)
= tanA (1 - 3 tan²A + 8) / (1 - 3 tan²A)
= 3tanA (3 - tan²A) / (1 - 3 tan²A)

tan2A = 2tanA / (1 - tan²A)

tan3A = tan(2A + A)
= (tan2A + tanA) / (1 - tan2A tanA)
= (2tanA + tanA(1 - tan²A)) / ((1 - tan²A) - 2tanA tanA)
= tanA (3 - tan²A)) / (1 - 3tan²A)

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