If A, B and C are interior angles of a triangle ABC, then what is \tan(\frac{B+C}{2})+\sin(\frac{B+C}{2})-\cot(\frac{A}{2})-\cos(\frac{A}{2}) equal to?

Explanation

In a triangle, A+B+C = 180^{\circ}, so \frac{B+C}{2} = 90^{\circ} - \frac{A}{2}. Therefore, \tan(\frac{B+C}{2}) = \cot(\frac{A}{2}) and \sin(\frac{B+C}{2}) = \cos(\frac{A}{2}). Substituting these makes the expression \cot(\frac{A}{2}) + \cos(\frac{A}{2}) - \cot(\frac{A}{2}) - \cos(\frac{A}{2}) = 0.

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