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
$$
 \cos \frac{a+b}{2} \sin \frac{a-b}{2} = \cos \frac{b+c}{2} \sin \frac{b-c}{2}
$$
 A: $$\tan\dfrac{c+a}2-\tan\dfrac{a+b}2$$
$$=\dfrac{\sin\dfrac{c+a-(a+b)}2}{\cos\dfrac{c+a}2\cos\dfrac{a+b}2}$$
$$=\dfrac{2\sin\dfrac{c-b}2\cos\dfrac{b+c}2}{2\cos\dfrac{c+a}2\cos\dfrac{a+b}2\cos\dfrac{b+c}2}$$
$$=\dfrac{\sin c-\sin b}{2\cos\dfrac{c+a}2\cos\dfrac{a+b}2\cos\dfrac{b+c}2}$$
Similarly, 
$$\tan\dfrac{b+c}2-\tan\dfrac{c+a}2=\dfrac{\sin b-\sin a}{2\cos\dfrac{c+a}2\cos\dfrac{a+b}2\cos\dfrac{b+c}2}$$
A: Let $a+b=2z$ etc.
$\implies2b=a+b+b+c-(c+a)=2(z+x-y)\implies b=z+x-y$
$$\sin(y+z-x)+\sin(x+y-z)=2\sin(z+x-y)$$
$$\sin(x+y-z)-\sin(z+x-y)=\sin(z+x-y)-\sin(y+z-x)$$
$$\sin(y-z)\cos x=\sin(x-y)\cos z$$
Expand using $\sin(A-B)$ formula and divide both sides by $\cos x\cos y\cos z$
A: 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.
