In a triangle ABC prove (b+c-a)(cot B/2+cotC/2)=2acot A/2

Dear student 

Cot(A/2) = s(s-a)/delta
Cot(B/2) = s(s-b)/delta
Cot(C/2) = s(s-c)/delta
LHS 
[Cot(A/2)+Cot(B/2)+Cot(C/2)]/Cot(A/2) = (a+b+c)/(b+c-a)
Put the values 
s = a+b+c /2 
you will get 
3s-2s /(s-a) = s/s-a
a+b+c /  (a+b+c-2a) = (a+b+c)/(b+c-a)
= RHS 
SO, 
(b+c-a)(cot B/2+cotC/2)=2acot A/2