(b-c)cot(A/2) + (c-a)coy(B/2) + (a-b)cot(C/2) = 0

L.H.S. =>

we know sine rule

a/sinA = b/sinB = c/sinC = k

a = ksinA , b = ksinB , c = ksinC

=> k[(sinB - sinC)cot(A/2) + (sinC - sinA)cot(B/2) + (sinA - sinB)cot(C/2)]

apply sinC - sinD = 2sin{(C-D)/2}cos{(C+D)/2}

=> k[2sin{(B-C)/2}cos{(B+C)/2}cot(A/2) + 2sin{(C-A)/2}cos{(C+A)/2}cot(B/2) + 2sin{(A-B/)2}cos{(A+B)/2}cot(C/2)]

in a triangle A+B+C = π

(A+B)/2 = π/2 - C/2 and so on for (B+C)/2 = π/2 - A/2 and (C+A)/2 = π/2 - B/2

=> k[2sin{(B-C)/2}sin(A/2)cot(A/2) + 2sin{(C-A)/2}sin(B/2)cot(B/2) + 2sin{(A-B/)2}sin(C/2)cot(C/2)]

=> k[2sin{(B-C)/2}cos(A/2) + 2sin{(C-A)/2}cos(B/2) + 2sin{(A-B/)2}cos(C/2)]

A+B+C = π

A/2 = π/2 - (B+C)/2 , B/2 = π/2 - (C+A)/2 , C/2 = π/2 - (A+B)/2

=> k[2sin{(B-C)/2}sin{(C+B)/2} + 2sin{(C-A)/2}sin{(A+C)/2} +

2sin{(A-B/)2}sin{(B+A)/2}]

apply cosC - cosD = 2sin{(C+D)/2}sin{(D-C)/2}

=> k[cosC - cosB + cosA - cosC + cosB - cosA]

=> 0 = R.H.S.

hence proved.

I think you can understand if any doubt please ask.