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)