simplify: Tan-1(1/3)+tan-1(1/2)
is there an algebraeic way to do this?
Yes.
Oh, now you want to see a way ......
There are many ways. Here's a simple but brutal one (watch out for falling Grand Pianos):
Let $\displaystyle \tan \alpha = \frac{1}{3}$ and $\displaystyle \tan \beta = \frac{1}{3}$.
Then $\displaystyle \tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{\frac{1}{3} + \frac{1}{2}}{1 - \left( \frac{1}{3} \right) \left( \frac{1}{2} \right)} = 1$.
Therefore $\displaystyle \alpha + \beta = \tan^{-1} \left( \frac{1}{3}\right) + \tan^{-1} \left( \frac{1}{2}\right) = \tan^{-1} (1) = \frac{\pi}{4}$.
Bet you didn't see that coming