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

Question

The problem is finding the sum of the numbers:

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

I've tried expressing $\frac{1}{2}$ and $\frac{1}{3}$ as tangent functions of some angles but I wasn't able to find a valid solution.

Any help would be much appreciated.

Answer 1

Hint:

$$\arctan(x)+\arctan(y)=\arctan\left[\dfrac{x+y}{1-xy} \right]$$

Answer 2

Well, since$$\tan\left(\arctan\frac12+\arctan\frac13\right)=\frac{\frac12+\frac13}{1-\frac12\times\frac13}=1\ldots$$

Answer 3

Let $x=\arctan\frac{1}{2}$ and $y=\arctan\frac{1}{3}$. Then $x,y\in[0,\frac{\pi}{4}]$ and hence $x+y\in[0,\frac{\pi}{2}]$.

$$\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\times\frac{1}{3}}=1$$

$x+y=\frac{\pi}{4}$