# In any ΔABC, prove that

Question:

In any ΔABC, prove that

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

Solution:

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

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

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

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

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

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

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