If `x , ya n dz` are the distances of incenter from the vertices of the triangle `A B C` , respectively, then prove that `(a b c)/(x y z)=cot(A/2)cot(B/2)cot(C/2)`

Explanation

`x = r cosec.(A)/(2) and a = r(cot.(B)/(2) + cot.(C)/(2))`
`rArr (a)/(x) (cot.(B)/(2) + cot.(C)/(2)) sin.(A)/(2) = (sin.(A)/(2) cos.(A)/(2))/(sin.(B)/(2) sin.(C)/(2))`
Similarly, `(b)/(y) = (sin.(B)/(2) cos.(B)/(2))/(sin.(A)/(2) sin.(C)/(2)) and (c)/(z) = (sin.(C)/(2) cos.(C)/(2))/(sin.(A)/(2) sin.(B)/(2))`
`:. (abc)/(xyz) = (cos.(A)/(2) cos.(B)/(2) cos.(C)/(2))/(sin.(A)/(2) sin.(B)/(2) sin.(C)/(2))`
`= cot.(A)/(2) cot.(B)/(2) cot.(C)/(2)`