For any triangle ABC prove that Sin B-C/2=B-C/A cos A/

Dear Student,
Please find below the solution to your problem.

Let a/sin A = b/sin B = c/sinC = k
Then,
a = k sinA, b = k sinB, c = k sinC
RHS
b-c/a cos A/2
= (ksinB - ksinC/k sinA )cos A/2
= {[2 cos B+C/2 Sin B-C/2]/ Sin A} cos A/2
=[2* Sin B-C/2 cos (pi/2-A/2) .cos A/2] / sinA
= [SinA* sinB-C/2] / sinA
= Sin B-C/2
= LHS
Hence proved

Thanks and Regards