Q : If tanA, tanB, tanC are the solutions of the equation x3 – k2 x2 – 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 = k2 and tanA tanB tanC = -2k – 1
In a ΔABC
tanA + tanB + tanC = tanA tanB tanC
⇒ k2 = -2k-1 ⇒ k = -1
⇒ tanA + tanB + tanC = tanA tanB tanC = 1
which is not possible.
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