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Question

If $$A,\,B,\,C$$ are angles of $$\Delta ABC$$ and $$tan\,A\,tan\,C=3,\,tan\,B\,tan\,C=6$$, 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 $$tan\,A\,tan\,B=2$$
$$\tan\,A+\tan\,B+\tan\,C=\tan\,A\,tan\,B\,\tan\,C=3\,\tan\,B=6\,\tan\,A$$
$$\therefore\;3\,\tan\,A+\tan\,C=6\,\tan\,A\,\tan\,C=3\,\tan\,A$$
$$\therefore\;\tan\,A+2\,\tan\,A+3\,\tan\,A=\tan\,A\cdot 2\,tan\,A\cdot 3\,\tan\,A$$
i.e. $$\tan\,A=\tan^3\,A$$
i.e. $$\tan^2\,A=1$$ i.e. $$\tan\,A=1$$
$$\therefore\;A=\displaystyle\frac{\pi}{4}$$
$$\therefore\;\tan\,A\,\tan\,B=2\,\tan^2\,A=2$$

Mathematics

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