Hello all. I have a trig problem here that I am not necessarily stuck on, but just wanted to confirm results with someone else.

Using the proof:

$\displaystyle \frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\b eta}$

Where:

$\displaystyle \tan\alpha = \frac{3\sqrt5}{5}$

$\displaystyle \tan\beta = \frac{-\sqrt15}{4}$

For:

$\displaystyle \tan(\alpha+\beta)$

What do you get?