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If A+B+C=0, then
The value of $|\begin{matrix} A & \tan A & c o r A \cot B \\ B & \tan B & \cot B \cot C \\ C & \tan C & \cot C \cot A \end{matrix}| = |\begin{matrix} λ & \tan A \tan B \tan C & 1 \\ - A & \tan B + \tan C & \cot C (\cot B + \cot A) \\ C & \tan C & \cot C \cot A \end{matrix}|$

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detailed solution

Correct option is A

A+B+C=0⇒tanA+tanB+tanC=tanAtanBtanC and cotAcotB+cotBcotC+cotCcotA=1Now, AtanAcotAcotBBtanBcotBcotCCtanCcotCcotA R1→R1+R2+R3, R2→R2+R3 =A+B+CtanA+tanB+tanCcotAcotB+cotBcotC+cotCcotAB+CtanB+tanCcotBcotC+cotCcotACtanCcotCcotA =0tanAtanBtanC1-AtanB+tanCcotCcotB+cotACtanCcotCcotA⇒λ=0

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