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Question

If tanB=2sinAsinCsin(A+C) then
tanA,tanB,tanC are in

A
A.P
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B
G,P
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C
H.P
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D
AGP
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Solution

The correct option is B H.P
tanB=2sinAsinCsin(A+C)sinBcosB=2sinAsinCsinAcosC+cosAsinCsinAsinBcosC+cosAsinBsinC=2sinAcosBsinC
Dividing both sides by sinAsinBsinC, we get
sinAsinBcosCsinAsinBsinC+cosAsinBsinCsinAsinBsinC=2sinAcosBsinCsinAsinBsinCcosCsinC+cosAsinA=2cosBsinB1tanC+1tanA=2tanB
Hence tanA,tanB,tanC are in H.P

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