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

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