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Q.

If A+B+C=0, thenThe value of AtanAcorAcotBBtanBcotBcotCCtanCcotCcotA=λtanAtanBtanC1−AtanB+tanCcotCcotB+cotACtanCcotCcotA

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a

0

b

1

c

2

d

-1

answer is A.

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Detailed Solution

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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