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
$$\tan\left(x+y\right)\:=\:\dfrac{\tan x\:+\tan y}{1-\left(\tan x\right)\left(\tan y\right)}$$

Let $$\arctan\left(\dfrac{1}{2}\right)+\arctan\left(\dfrac{1}{3}\right)=u $$
Take tangents on both sides using hint given.
$$ \dfrac{1/2 +1/3}{1- {\dfrac{1} {6}}} = tan(u), $$
$$  tan(u) =1$$
$$ u = \pi/4 $$
