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Question

If sinAsinBsinC=p and cosAcosBcosC=q, where A,B,C are the angles of a ΔABC, then tanAtanB+tanBtanC+tanCtanA is equal to

A
pq
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B
q+1q
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C
p+1p
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D
q+1p
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Solution

The correct option is B q+1q

For triangle ABC, A+B+C=π

1=cos(A+B+C)


1=cos((A+B)+C)


1=cos(A+B)cosCsin(A+B)sinC


1=(cosAcosBsinAsinB)cosC((sinAcosB+cosAsinB)sinC)


1=cosAcosBcosCsinAsinBcosCsinAcosBsinCcosAsinBsinC


1=cosAcosBcosC(1sinAsinBcosAcosBsinAsinCcosAcosCsinCsinBcosCcosB)


1=q(1tanAtanBtanAtanCtanBtanC)


1q=1(tanAtanB+tanAtanC+tanBtanC)


tanAtanB+tanAtanC+tanBtanC=1+1q


tanAtanB+tanAtanC+tanBtanC=q+1q


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