prove that a(cosC-cosB)=2(b-c)cos^2(A/2)

Hello student, Please find the answer to your question below

a/ 2(b-c) = cos^2(A/2)/ (cosC-cosB)
lhs = a / 2 (b – c )
Now use sine rule
a/ sinA = b/ sinB = c/ sinC = k
So, 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)
=cos²(A/2)/ Cos(A/2) cos(B-C/2)
= cos²(A/2)/ Cos(90 – (B+C)/2) cos(B-C/2)
=2 cos²(A/2)/ 2 sin((B+C)/2) cos(B-C/2)
= 2 cos²(A/2)/ 2 sin((B+C)/2) cos(B-C/2)
=cos²(A/2)/ (cosC-cosB)