Find the value of the sum $$\arctan\left(\dfrac{1}{2}\right)+\arctan\left(\dfrac{1}{3}\right)$$
We were also given a hint of using the trigonometric identity of $\tan(x + y)$
Hint
$$\tan\left(x+y\right)\:=\:\dfrac{\tan x\:+\tan y}{1-\left(\tan x\right)\left(\tan y\right)}$$

Yep, sum of tangents identity is the way to go. Write $x=arctan\frac{1}{2}$ and $y=arctan\frac{1}{3}$. 
Then $tan(arctan\frac{1}{2}+arctan\frac{1}{3})=\frac{tan(arctan\frac{1}{2})+tan(arctan\frac{1}{3})}{1-(tan(arctan\frac{1}{2}))(tan(arctan\frac{1}{3}))}$
This is equal to $\frac{\frac{1}{2}+\frac{1}{3}}{1-(\frac{1}{2})(\frac{1}{3})}=\frac{\frac{5}{6}}{\frac{5}{6}}=1$
Thus $tan(x+y)=1$ and $x+y=\boxed{\frac{\pi}{4}}$
