If A+B+C=πthen the value of tanA+tanB+tanCtanA.tanB.tanC+cotAtanB+cotBtanC+cotCtanA
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a
1
b
2
c
3
d
4
answer is B.
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Detailed Solution
A+B+C=π⇒ tanA+tanB+tanC=tanAtanBtanC and cotAcotB+cotBcotC+cotCcotA=1⇒cotAtanB+cotBtanC+cotCtanA=1 ∴tanA+tanB+tanCtanAtanBtanC+cotAtanB+cotBtanC+cotCtanA=1+1=2