In any triangle ABC,Prove that a(cosC-cosB)=2(b-c)cos²A/2

(A/2)/ (cosC-cosB)²(A/2)/ 2 sin((B+C)/2) cos(B-C/2)=cos²(A/2)/ 2 sin((B+C)/2) cos(B-C/2)= 2 cos²(A/2)/ Cos(90 – (B+C)/2) cos(B-C/2)=2 cos²(A/2)/ Cos(A/2) cos(B-C/2)= cos²a/ 2(b-c) = cos^2(A/2)/ (cosC-cosB)lhs = a / 2 (b – c )Now use sine rulea/ sinA = b/ sinB = c/ sinC = kSo, lhs = a / 2 (b – c )= sinA / 2(sinB -sinC)=SinA /2* 2 cos(B+C/2) cos (B-C/2)= sinA / 4 cos (pi/2 – A/2) cos (B-C/2)=2 sin(A/2) cos (A/2) / 4 sin(A/2) cos (B-C/2)= cos (A/2) / cos(B-C/2)=cos