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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
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