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Question

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 B tanAtanB=2
tanA+tanB+tanC=tanAtanBtanC=3tanB=6tanA
3tanA+tanC=6tanAtanC=3tanA
tanA+2tanA+3tanA=tanA2tanA3tanA
i.e. tanA=tan3A
i.e. tan2A=1 i.e. tanA=1
A=π4
tanAtanB=2tan2A=2

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