I'm practicing trig and so I'm working through some exercises. I'm having some trouble figuring this one out:

$\displaystyle tan(\alpha + \beta) = \frac{tan\, \alpha + tan\, \beta}{1 - tan\, \alpha\, tan\, \beta}$

First I've rewritten $\displaystyle tan(\alpha + \beta)$ as $\displaystyle \frac{sin(\alpha + \beta) }{cos(\alpha + \beta)}$ which equates to:

$\displaystyle \frac{sin\, \alpha\, cos\, \beta + \cos\,\alpha\,sin\beta}{cos\,\alpha\,cos\beta - sin\,\alpha\,sin\,\beta}$.

Then I'm kind of stuck, so I tried going the other way around:

$\displaystyle \frac{tan\,\alpha + tan\,\beta}{1 - tan\,\alpha \, tan\, \beta}$ =

$\displaystyle \frac{\frac{sin\,\alpha}{cos\,\alpha}+ \frac{sin\alpha\,\beta }{cos\,\beta}}{1 - \frac{sin\, \alpha}{cos\, \alpha}\frac{sin\, \beta}{cos\, \beta}}$

Then I multiply the 1/1 in the denominator by cos a cos b, to make them equal, to get rid of the 1, then I rewrite it according to $\displaystyle \frac{\frac{a}{b}}{\frac{c}{d}}$ = $\displaystyle \frac{ad}{bc}$ which leaves me with a lot of sin and cos, but stuck as well. Some help would be nice