We transform the equation $$(a+b)\sin\frac{\gamma}{2}=c\cos\left(\frac{\alpha -\beta}{2}\right)$$ to $$a\cos\left(\frac{\alpha}{2}+\frac{\beta}{2}\right) +b\cos\left(\frac{\alpha}{2}+\frac{\beta}{2}\right)= c\cos\frac{\alpha}{2}\cos\frac{\beta}{2} +c\sin\frac{\alpha}{2}\sin\frac{\beta}{2} ,$$ using the fact that $\gamma=\pi-\alpha-\beta$ The left hand side can be further expanded, so that we get: $$a\left(\cos\frac{\alpha}{2}\cos\frac{\beta}{2}- \sin\frac{\alpha}{2}\sin\frac{\beta}{2}\right)+ b\left(\cos\frac{\alpha}{2}\cos\frac{\beta}{2}- \sin\frac{\alpha}{2}\sin\frac{\beta}{2}\right)= c\cos\frac{\alpha}{2}\cos\frac{\beta}{2} +c\sin\frac{\alpha}{2}\sin\frac{\beta}{2}.$$ Collecting terms we get: $$(a+b-c)\cos\frac{\alpha}{2}\cos\frac{\beta}{2}- (a+b+c)\sin\frac{\alpha}{2}\sin\frac{\beta}{2}=0.$$ Using $s:=\frac{a+b+c}{2}$ and using the equations \begin{eqnarray*} \sin\frac{\alpha}{2}&=&\sqrt{\frac{(s-b)(s-c)}{bc}}\\ \cos\frac{\beta}{2}&=&\sqrt{\frac{s(s-a)}{bc}} \end{eqnarray*}we get: $$2\frac{s(s-c)}{c}\sqrt{\frac{(s-a)(s-b)}{ab}}- 2\frac{s(s-c)}{c}\sqrt{\frac{(s-a)(s-b))}{ab}}=0,$$ which is obviously true. So we can prove the first equation by going backwards. The second equation can be proved in quite the same way.