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Prove the identity… (COS B- COS C) / (SIN B +SIN C) =-TAN ((B-C) /2)?
Okay.
Now, we'll use the identity we just mentioned. Note that B+C = (B-C) + 2C. So, we can write:

[COS B - COS C] / [SIN (B+C)]
= [COS B - COS C] / [SIN (B-C+2C)]
= [COS B - COS C] / [SIN (B-C) COS 2C + SIN 2C COS(B-C)]