In triangle ABC, prove the following:

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

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

$\mathrm{LHS}=(a-b) \cos \frac{C}{2}$

$=k(\sin A-\sin B) \cos \frac{C}{2} \quad($ using $(1))$

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

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

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

$=k \sin \left(\frac{A-B}{2}\right) \sin (A+B) \quad\left[\because 2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A+B}{2}\right)=\sin (A+B)\right]$

$=k \sin \frac{A-B}{2} \sin (\pi-\mathrm{C}) \quad[\because A+B+C=\pi]$

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

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

Hence proved.