Question

In ` A B C ,` if `cotA+cotB+cotC=0` then find the value of `cosAcos BcosC`

Answer 1

`cotA+cotB+cotC = 0`
`=>cosA/sinA +cosB/sinB = -cosC/sinC`
`=>(cosAsinB+sinAcosB)/(sinAsinB) = -cosC/sinC`
`=>(sin(A+B))/(sinAsinB) = -cosC/sinC`
`=>(sin(pi-C))/(sinAsinB) = -cosC/sinC`
`=>(sinC)/(sinAsinB) = -cosC/sinC`
`=>(sin^2C)/(sinAsinB) = -cosC->(1)`
Similarly,
`=>(sin^2B)/(sinAsinC) = -cosB->(2)`
`=>(sin^2A)/(sinBsinC) = -cosA->(3)`
Multiplying (1),(2) and (3),
`1 = -cosAcosBcosC`
`:. cosAcosBcosC= -1.`