In any ΔABC, prove that

Need to prove: $\frac{(b+c)}{a} \cdot \cos \frac{(B+C)}{2}=\cos \frac{(B-C)}{2}$

We know that, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R$ where $R$ is the circumradius.

Therefore,

$a=2 R \sin A \cdots$ (a)

Similarly, $b=2 R \sin B$ and $c=2 R \sin C$

Now, $\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}=\frac{2 \mathrm{R} \sin \mathrm{A}}{2 \mathrm{R} \sin \mathrm{B}+2 \mathrm{R} \sin \mathrm{C}}=\frac{\sin \mathrm{A}}{\sin \mathrm{B}+\sin \mathrm{C}}$

$\Rightarrow \frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}=\frac{2 \sin \frac{\mathrm{A}}{2} \cos \frac{\mathrm{A}}{2}}{2 \sin \frac{\mathrm{B}+\mathrm{C}}{2} \cos \frac{\mathrm{B}-\mathrm{C}}{2}}$

$\Rightarrow \frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}=\frac{\sin \frac{\mathrm{A}}{2} \cos \frac{\mathrm{A}}{2}}{\sin \left(\frac{\pi}{2}-\frac{\mathrm{A}}{2}\right) \cos \frac{\mathrm{B}-\mathrm{C}}{2}}$

$\Rightarrow \frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}=\frac{\sin \frac{\mathrm{A}}{2} \cos \frac{\mathrm{A}}{2}}{\cos \frac{\mathrm{A}}{2} \cos \frac{\mathrm{B}-\mathrm{C}}{2}}$

$\Rightarrow \frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}=\frac{\cos \left(\frac{\pi-\frac{\mathrm{A}}{2}}{2}\right)}{\cos \left(\frac{\mathrm{B}-\mathrm{c}}{2}\right)}$

$\Rightarrow \frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}=\frac{\cos \left(\frac{\pi-\mathrm{A}}{2}\right)}{\cos \left(\frac{\mathrm{B}-\mathrm{c}}{2}\right)}$

$\Rightarrow \frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}=\frac{\cos \left(\frac{\mathrm{B}+\mathrm{c}}{2}\right)}{\cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)}$

$\Rightarrow \frac{\mathrm{b}+\mathrm{c}}{\mathrm{a}} \cos \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)=\cos \frac{\mathrm{B}-\mathrm{C}}{2}$ [Proved]