A very general (but extremely useful) approach is by noting the following
$$\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$$
$$\cos(x) = \frac{e^{ix} + e^{-ix}}{2}$$
and since $\tan(x) = \frac{\sin(x)}{\cos(x)}$
$$\tan(x) = \frac{1}{i}\frac{e^{ix} - e^{-ix}}{e^{ix} + e^{-ix}} = -i \frac{e^{ix} - e^{-ix}}{e^{ix} + e^{-ix}} $$
We note here that $i$ is the imaginary constant (basically its a number such that $i^2 = -1$)
I will leave you to go ahead and verify these formulas work for every trig Identity you have already memorized and more information is underneath:
http://en.wikipedia.org/wiki/Euler%27s_formula
So we wish to prove:
$$\frac{\sin(A+B)}{\sin(A-B)} = \frac{\tan(A) + \tan(B)}{\tan(A) - \tan(B)} $$
We prepare the left side (noting that both fractions take form $\frac{A}{2i}$ and therefore we can drop the $2i$ denominators
$$\frac{\sin(A+B)}{\sin(A-B)} = \frac{e^{i(A+B)} - e^{-i(A+B)}}{e^{i(A-B)} - e^{-i(A-B)}}$$
So because I'm slightly lazy (and for the sake of giving you something to practice) you need to do the exact same job with the tan(x) expression where each instance of x becomes A or B depending on whats being evaluated.
Now the goal is to systematically simplify both expressions by transforming expressions of the form $e^{-k}$ to $\frac{1}{e^k}$
Followed by taking sums and giving them common denominators $\frac{A}{C} + \frac{B}{D} = \frac{AD + BC}{CD}$
And dividing out common factors.
Its a tedious process but once done. Both expression will look exactly the same... Well you don't have to take my word for it, do it yourself and prove that it works ;)
'
at the end of the denominator in your equation is incorrect. This is simply a comma,
in the original printed formula. $\endgroup$ – Pixel Feb 8 '14 at 17:28