Explanation of inverse trig euation solution The equation is:
$\arctan 3 + 2\arctan2 = \pi + arccot 3$
They go on and assign $\arctan 3 = \theta$ and $\arctan2 = \phi$. Therefore, $\frac{\pi}{4}< \theta < \frac{\pi}{2}$, same for $\phi$. Which I follow.
Then they use $\tan(A+B)$ formula to show that that $\tan(\theta+2\phi)=\frac{1}{3}$. Which implies that $\arctan \frac{1}{3} = \theta + 2\phi$, but apparently not so. As it implies $\pi + \arctan \frac{1}{3} = \theta + 2\phi$.
And there I am lost. I can see that when you combine the inequalities you get $$\pi+\frac{\pi}{4} < \theta + 2\phi < \pi + \frac{\pi}{2}$$
And then they go on to show that $$\theta + 2\phi = \pi + \arctan \frac{1}{3} = \pi + arccot 3$$
I do not understand where that $\pi$ is coming from there...

First of all, $\arctan x+\text{arccot}x=\dfrac\pi2$  (Proof)
So, the problem reduces to $\arctan 2+\arctan3=\dfrac{3\pi}4$
Now from this or Ex$\#5$  of Page $\#276$ of this,
  $$\tan^{-1}x+\tan^{-1}y=\begin{cases} \tan^{-1}\dfrac{x+y}{1-xy} &\mbox{if } xy<1 \\\pi+ \tan^{-1}\dfrac{x+y}{1-xy} & \mbox{if } xy> 1. \end{cases} $$ 
Finally,   the principal value of $\displaystyle\tan$ lies $\displaystyle\in\left[-\frac\pi2,\frac\pi2\right]$
