If A,B,C are angles of ΔABC and tanAtanC=3,tanBtanC=6, then
A
A=π4
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B
tanAtanB=2
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C
tanAtanC=32
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D
tan3A+tan3B=−tan3C+3tanAtanBtanC=0
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Solution
The correct option is BtanAtanB=2 tanA+tanB+tanC=tanAtanBtanC=3tanB=6tanA ∴3tanA+tanC=6tanAtanC=3tanA ∴tanA+2tanA+3tanA=tanA⋅2tanA⋅3tanA i.e. tanA=tan3A i.e. tan2A=1 i.e. tanA=1 ∴A=π4 ∴tanAtanB=2tan2A=2