Proving $\tan\left( \frac {a+b}{2}\right) = \frac{\sin a+\sin b} {\cos a + \cos b } $ 
Prove $$\tan\left( \frac {a+b}{2}\right) = \frac{\sin a+\sin b} {\cos a + \cos b } $$

Can someone help? I separated $\tan$ and did double-angle, but I just went into a circle and couldn't get the trig functions without the halves.

Use the formulas
$$\sin a+\sin b=2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right),$$
$$\cos a+\cos b=2\cos\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right).$$
Therefore, we have
$$\frac{\sin a+\sin b}{\cos a+\cos b}=\frac{\sin\displaystyle\left(\frac{a+b}{2}\right)}{\cos\displaystyle\left(\frac{a+b}{2}\right)}=\tan\left(\frac{a+b}{2}\right).$$
