If A+B+C = 0, and tan A + tan B + tan C = 0, find the value of \(\frac{tan^2A}{tanBtanC}+\frac{tan^2B}{tanAtanC}+\frac{tan^2C}{tanAtanB}\)

1 2 3 4

3

If A + B + C = 0 A3 + B3 + C3 = 3ABC ⇒ \(\frac{tan^3A + tan^3B+tan^3C}{tanAtanBtanC}\) ⇒ \(\frac{3tanAtanBtanC}{tanAtanBtanC}\) = 3