In Delta ABC,(b - c)cot frac{A}{2} + (c - a)c | vedclass

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(A) Let $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$. Substituting these into the expression: $(b - c) \cot \frac{A}{2} = 2R(\sin B - \sin C) \

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$0$

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(A) Let $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$. Substituting these into the expression: $(b - c) \cot \frac{A}{2} = 2R(\sin B - \sin C) \cot \frac{A}{2}$ $= 2R \left( 2 \cos \frac{B+C}{2} \sin \frac{B-C}{2} \right) \cot \frac{A}{2}$ Since $\frac{B+C}{2} = 90^\circ - \frac{A}{2}$,$\cos \frac{B+C}{2} = \sin \frac{A}{2}$. $= 2R \left( 2 \sin \frac{A}{2} \sin \frac{B-C}{2} \right) \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}} = 4R \sin \frac{B-C}{2} \cos \frac{A}{2}$ $= 4R \sin \frac{B-C}{2} \sin \frac{B+C}{2} = 2R(\cos(B-C) - \cos(B+C)) = 2R(\cos(B-C) + \cos A)$. Summing this cyclically for all terms results in $0$.

In $\Delta ABC$,$(b - c)\cot \frac{A}{2} + (c - a)\cot \frac{B}{2} + (a - b)\cot \frac{C}{2}$ is equal to

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