Simplify:

$\displaystyle 3arctan(2) - arctan(2/11)$

I'd know how to do it if it weren't for the 3 infront of arctan(2).

EDIT: I guess I could do it like this:

$\displaystyle arctan(2) + arctan(2) + arctan(2) - arctan(2/11)$

That seems a bit over the top though. I'm still looking for an easier solution.

Also, I've never encountered three arctan terms before:

$\displaystyle arctan(2) + arctan(3) + arctan(4)$

I'm guessing I need to first simplify $\displaystyle arctan(2) + arctan(3)$, and then use that to get $\displaystyle (arctan(2) + arctan(3)) + arctan(4)$ by using the $\displaystyle tan(u+v)$ rule again. And of course make sure it's in the range of $\displaystyle \pi/2<arctan(2)+arctan(3)+arctan(4)<3\pi/2$