If `triangleABC`, if `cot A+cotB+cotC=0` then find the value of `cos A cos B cos C`.

Explanation

`cot A+cot B+cotC=0`
`rArr cot B+cot C=-cotA`
`rArr(sin B cos C+sinC cos B)/(sin B sin C)=(-cos A)/(sin A)`
`rArr(sin(B+C))/(sin BsinC)=(-cos A)/(sin A)`
`rArr (sin^(2)A)/(sin B sin C)=-cos A(because sin(B+C)=sin(pi-A)`
`=sinA)`
Similarly `(sin^(2)B)/(sin A sin C)=-cos B`
and `(sin^(2)C)/(sin A sin B)=-cos C`
Multiplying eqs. i, ii and iii, we get `cos A cos B cos C=-1`