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$\tan A, \tan B, \tan C$ are the roots of the cubic equation $x^{3} - 7 x^{2} + 11 x - 7 = 0$ then $A + B + C c a n b e$

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Correct option is B

We have tanA+tanB+tanC=7 and tanA·tanB·tanC=7⇒tanA+tanB+tanC= tanA·tanB·tanC⇒A+B+C=nπ⇒A+B+C=π

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tanA,tanB,tanCare the roots of the cubic equation x3−7x2+11x−7=0 then A+B+C can be

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$\tan A, \tan B, \tan C$ are the roots of the cubic equation $x^{3} - 7 x^{2} + 11 x - 7 = 0$ then $A + B + C c a n b e$