Show that Tan(A+B+C) = (tanA + tanB + tanC – tanAtanBtanC) ÷ (1-tanAtanB-tanBtanC-tanAtanC). Hence deduce that Tan3A = (3tanA-tan³A)/(1-3tan³A)

Trigonometry analytics

  1. Given that tan(A+B) = (tanA + tanB)/(1-tanAtanB). Show that Tan(A+B+C) = (tanA + tanB + tanC – tanAtanBtanC) ÷ (1-tanAtanB-tanBtanC-tanAtanC).
    Hence deduce that Tan3A = (3tanA-tan³A)/(1-3tan³A)
    Solvings
    Tan (A+B+C) = Tan[(A+B)+C]
    Let Tan(A+B) = Tanx
    Tan(X+C) = (Tanx + TanC)÷ (1-Tanxtanc)

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