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# In triangle ABC if cosA cosB cosB=1/3 then find the value of tanAtanB+tanBtanC+tanCtanA

Grade:11

## 2 Answers

Mayur Reddy
23 Points
5 years ago

Given cosAcosBcosC=1/3

then tanAtanB +tanB tanC+ tanC tanA=sinAsinBcosC +sinBsinCcosA +sinCsinAcosB/cosAcosBcosC

tanAtanB +tanB tanC+ tanC tanA=(sinA sinB cosC +sinB sinC cosA +sinC sinA cosB)/cosA cosB cosC

=3 (sinA sinB cosC +sinC [sinB cosA + sinA cosB])

=3 (sinA sinB cosC +sinC [sin(A+B)])

=3 (sinA sinB cosC +sin2C )

=3 (sinA sinB cosC +1-cos2C )

=3 (1+cosC{-cosC+sinA sinB } )

=3 (1+cosC cosA cosB  )

=3 (1+1/3 )

=3 (4/3)

=4

Vedit
20 Points
3 years ago
tanAtanB+ tanBtanC+ tanCtanA=1+ secAsecBsecC
SsecAsecBsecC=3
Therefore,tanAtanB+ tanBtanC+ tanCtanA=1+3 =4

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