simplify: Tan-1(1/3)+tan-1(1/2)

is there an algebraeic way to do this?

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- Jan 27th 2008, 09:16 PMJOhkonutarctan(1/3)+arctan(1/2)
simplify: Tan-1(1/3)+tan-1(1/2)

is there an algebraeic way to do this? - Jan 27th 2008, 11:14 PMmr fantastic
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 ;) - Jan 27th 2008, 11:41 PMtopher0805
I think the easiest way is to just use the special triangles. Does that not work in this case?

- Jan 27th 2008, 11:45 PMmr fantastic
- Jan 27th 2008, 11:49 PMtopher0805
Oh, I see. 2 is never an adjacent side. Disregard my last post.