If tan⁡A,tan⁡B and tan⁡C are the roots of equation x3−αx2−β=0 and 1+tan2⁡A1+tan2⁡B1+tan2⁡C=k−1, then k is

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If $\tan A, \tan B and \tan C$ are the roots of equation $x^{3} - α x^{2} - β = 0$ and $(1 + \tan^{2} A) (1 + \tan^{2} B) (1 + \tan^{2} C) = k - 1,$ then k is

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$(α - β)^{2}$

$1 + (α - β)^{2}$

$2 + (α - β)^{2}$

None of the above

answer is C.

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Detailed Solution

We have, $\tan A, \tan B, \tan C are roots of x^{3} - α x^{2} - β = 0,$ then

$Σtan A = \tan A + \tan B + \tan C = α_{,}$

$Σtan, \hat{ian} B = \tan A \tan B + \tan B \tan C + \tan C \tan A = 0$

and $\tan A \tan B \tan C = β$

Now, $(1 + \tan^{2} A) (1 + \tan^{2} B) (1 + \tan^{2} C)$

$\begin{array}{l} = 1 + {Σtan}^{2} A + {Σtan}^{2} A \tan^{2} B + \tan^{2} A \tan^{2} B \tan^{2} C \\ = 1 + (Σtan A)^{2} - 2 Σtan A \tan B + (Σtan A \tan B)^{2} - 2 \tan A \tan B \tan C Σtan A + (\tan A \tan B \tan C)^{2} \end{array}$