If A+B+C =pie,then prove that ∑(tanA÷tanBtanC) = ∑(tanA) -2∑(cotA).

A+B+C = 180 ;

A+B = 180 – C ;

/* take tangent on both sides and simplifiy

Tan A + Tan B + Tan C = (Tan A)*(Tan B)*(Tan C)

take LCM of LHS in expanded form */

(Tan ²A +Tan²B+Tan²C) ./ ((Tan A)*(Tan B)*(Tan C))

simplify using (a^2 + b^2 + c^2) = (a+b+c)^2 – 2(ab + bc + ca);

you will get RHS