$$\arctan \frac{1}{3} + \arctan \frac{1}{2} = \arctan 1$$
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 and here.
$$\arctan \frac{1}{3} + \arctan \frac{1}{2} = \arctan 1$$
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
$$ \arctan \frac{a}{b} + \arctan \frac{b-a}{b+a} = \arctan 1, \text{ for } a,b \in \mathbb{N}, a \leq b $$
$$\arctan \frac{1}{3} + \arctan \frac{1}{2} = \arctan 1$$
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 and here.
$$\arctan \frac{1}{3} + \arctan \frac{1}{2} = \arctan 1$$
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
$$ \arctan \frac{a}{b} + \arctan \frac{b-a}{b+a} = \arctan 1, \text{ for } a,b \in \mathbb{N}, a \leq b $$
