Giải thích các bước giải:
b.$\dfrac{\cos B+ \cos C}{\sin B+ \sin C}=\dfrac{2\cos \dfrac{B+C}{2}.\cos \dfrac{B-C}{2}}{2\sin\dfrac{B+C}{2}.\cos \dfrac{B-C}{2}}=\dfrac{\sin \dfrac{A}{2}}{\cos \dfrac{A}{2}}$
$\rightarrow \dfrac{\sin \dfrac{A}{2}}{\cos \dfrac{A}{2}}=\sin A=2\sin \dfrac{A}{2}. \cos \dfrac{A}{2}$
$\rightarrow 2 \cos ^2\dfrac{A}{2}-1=0$
$\rightarrow cosA=0$
$\rightarrow A=\dfrac{\pi}{2}$
$\rightarrow\Delta BAC$ vuông tại A