In a triangle ABC, $\text{∑}\frac{\cos \left(\mathrm{B}-\mathrm{C}\right)}{\sin \text{B sinC}}\text{=}$

$\begin{array}{rcl} & & \sum \frac{\cos (\mathrm{B}-\mathrm{C})}{\sin \mathrm{B} \sin \mathrm{C}}\\ & =& \sum \frac{\cos \mathrm{B} \cos \mathrm{C}+\sin \mathrm{B} \sin \mathrm{C}}{\sin \mathrm{B} \sin \mathrm{C}}\\ & =& \sum (\cot \mathrm{B} \cot \mathrm{C}+1)\\ & =& \cot \mathrm{B} \cot \mathrm{C}+1+\cot \mathrm{C} \cot \mathrm{A}+1+\cot \mathrm{A} \cot \mathrm{B}+1\\ & =& \cot \mathrm{A} \cot \mathrm{B}+\cot \mathrm{B} \cot \mathrm{C}+\cot \mathrm{C} \cot \mathrm{A}+3\\ & =& 1+3=4\end{array}$