$$\arctan \frac{1}{3} + \arctan \frac{1}{2} = \arctan 1$$

[![enter image description here][1]][1]


It's easy to generalize this to

$$ \arctan \frac{1}{n} + \arctan \frac{n-1}{n+1} = \arctan 1, \text{ for } n \in \mathbb{N}$$

which can further be generalized to

$$ \arctan \frac{a}{b} + \arctan \frac{b-a}{b+a} = \arctan 1, \text{ for } a,b \in \mathbb{N}, a \leq b $$

Edit: A similar result relating Fibonacci numbers to arctangents can be found [here][2] and [here][3].


  [1]: https://i.sstatic.net/sf56R.png
  [2]: http://www.futilitycloset.com/2016/06/05/a-guest-appearance/
  [3]: https://www.maa.org/sites/default/files/Fibonacci_Numbers55908.pdf