If $ \sin a +\sin c =2 \sin b $, show that $ \tan\frac{a+b}{2}+\tan\frac{b+c}{2} = 2 \tan\frac{c+a}{2}$ Found the question in the textbook.
I tried many methods of manipulating the identity to be proved but I did not even see a clue of using the given condition(of different forms like
$$
\sin\frac{a+c}{2} \cos\frac{a-c}{2} = \sin \frac{b}{2} \cos \frac{b}{2}
$$
and
$$
 \cos \frac{a+b}{2} \sin \frac{a-b}{2} = \cos \frac{b+c}{2} \sin \frac{b-c}{2}
$$

We need to prove that
$$\frac{\sin\left(\frac{a+b}{2}-\frac{c+a}{2}\right)}{\cos\frac{a+b}{2}\cos\frac{a+c}{2}}=\frac{\sin\left(\frac{c+a}{2}-\frac{b+c}{2}\right)}{\cos\frac{c+a}{2}\cos\frac{b+c}{2}}$$ or
$$\frac{\sin\frac{b-c}{2}}{\cos\frac{a+b}{2}}=\frac{\sin\frac{a-b}{2}}{\cos\frac{b+c}{2}}$$ or
$$\sin{b}-\sin{c}=\sin{a}-\sin{b},$$ which is given.
