$-$ Como temos um triângulo: \begin{matrix} A &+& B &+& C &=& 180^{\circ} \end{matrix}Da relação do enunciado, \begin{matrix} \tan{B} &=& \large{ \frac{\cos{(B-C)}}{\sin{(B+C)} \ + \ \sin{(C-B)}}} &=& \large{ \frac{\cos{(B)} \cos{(C)} \ + \ \sin{(B)}.\sin{(C)} }{ \sin{(B)}.\cos{(C)} \ + \ \sin{(C)}.\cos{(B)} \ + \ \sin{(C)}.\cos{(B)} \ - \ \sin{(B)}.\cos{(C)} }} \end{matrix}Continuando,\begin{matrix} \tan{B} &=& \large{ \frac{\cos{(B)} \cos{(C)} \ + \ \sin{(B)}.\sin{(C)} }{ 2\sin{(C)}.\cos{(B)} }} &=& \frac{1}{2} \ . \ (\cot{C} \ + \ \tan{B}) \end{matrix}Logo, \begin{matrix} \cot{C} = \tan{B} &\Rightarrow& {\large{\frac{\cos{C}}{\sin{C}} }}={\large{ \frac{\sin{B}}{\cos{B}}}} &\Rightarrow& \cos{(B)} \cos{(C)} \ - \ \sin{(B)}.\sin{(C)} = 0 \end{matrix}Portanto, \begin{matrix} \cos{(B+C)} = 0 &\Rightarrow& B+C = 90^{\circ} &\therefore& \fbox{$ A = 90^{\circ} $} \end{matrix}