Q : If tanA, tanB, tanC are the solutions of the equation x^{3} – k^{2} x^{2} – px + 2k + 1 = 0, then ΔABC is

(A) an isosceles triangle

(B) an equilateral triangle

(C) a right angled triangle

(D) none of these

Ans: (D)

Sol. tanA + tanB + tanC = k^{2} and tanA tanB tanC = -2k – 1

In a ΔABC

tanA + tanB + tanC = tanA tanB tanC

⇒ k^{2} = -2k-1 ⇒ k = -1

⇒ tanA + tanB + tanC = tanA tanB tanC = 1

which is not possible.