prove$\frac{  \sin a\vphantom{(}}{\sin b} +\frac{\cos a\vphantom{(}}{\cos b} = \frac{2\sin (a+b)}{\sin 2b}$ I've got this far but don't understand where the $2$ on the numerator comes from:
$$\dfrac{\sin a \cos b + \cos a \sin b}{\sin b \cos b}\overset{?}{=}\dfrac{\sin(a+b)}{\sin 2b}$$

$$\dfrac {\sin a}{\sin b}+\dfrac{\cos a}{\cos b}=\dfrac{\sin a\cdot\cos b+\cos a\cdot\sin b}{\sin b\cdot \cos b}$$
After getting this far, you need to observe that, $\boxed{\sin 2b=2 \cdot\sin b \cdot \sin b}$, you'll have to multiply the numerator and denominator by $2$, you'll have the following, 
$$\dfrac{2(\sin a\cdot\cos b+\cos a\cdot\sin b)}{2\cdot\sin b\cdot \cos b}=\dfrac{2\sin (a+b)}{\sin 2b}$$
Hope this helps.
