In triangle ABC, prove the following:

Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$         ...(1)

Then,

Consider the LHS of the equation $\frac{a+b}{c}=\frac{\cos \left(\frac{A-B}{2}\right)}{\sin \frac{C}{2}}$.

$\mathrm{LHS}=\frac{a+b}{c}$

$=\frac{k \sin A+k \sin B}{k \sin C} \quad($ using $(1))$

$=\frac{2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}{2 \sin \frac{C}{2} \cos \frac{C}{2}}$

$=\frac{\sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}{\sin \frac{C}{2} \cos \left(\frac{\pi-(A+B)}{2}\right)} \quad(\because A+B+C=\pi)$

$=\frac{\sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}{\sin \frac{C}{2} \sin \left(\frac{A+B}{2}\right)}$

$=\frac{\cos \left(\frac{A-B}{2}\right)}{\sin \frac{C}{2}}=\mathrm{RHS}$

Hence proved.