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Prove the identity… (COS B- COS C) / (SIN B +SIN C) =-TAN ((B-C) /2)?
Hmm, I can see where this is going.
Now, we can factor out the common factor of SIN (B-C) from the denominator, we get:

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