<s>[INST] 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
$$
 \cos \frac{a+b}{2} \sin \frac{a-b}{2} = \cos \frac{b+c}{2} \sin \frac{b-c}{2}
$$
 [/INST] $$\tan\dfrac{c+a}2-\tan\dfrac{a+b}2$$
$$=\dfrac{\sin\dfrac{c+a-(a+b)}2}{\cos\dfrac{c+a}2\cos\dfrac{a+b}2}$$
$$=\dfrac{2\sin\dfrac{c-b}2\cos\dfrac{b+c}2}{2\cos\dfrac{c+a}2\cos\dfrac{a+b}2\cos\dfrac{b+c}2}$$
$$=\dfrac{\sin c-\sin b}{2\cos\dfrac{c+a}2\cos\dfrac{a+b}2\cos\dfrac{b+c}2}$$
Similarly, 
$$\tan\dfrac{b+c}2-\tan\dfrac{c+a}2=\dfrac{\sin b-\sin a}{2\cos\dfrac{c+a}2\cos\dfrac{a+b}2\cos\dfrac{b+c}2}$$
</s>