Problem: Given that $0<\alpha<\frac{\pi}{2}$, $0<\beta<\frac{\pi}{2}$, $\tan{\alpha=2}$ and that $\tan{\beta=3},$ find $\alpha + \beta.$
Attempt: Given the restrictions on $\alpha$ and $\beta$, it follows that $0<\alpha+\beta<\pi.$ So depending on if $\alpha+\beta \ $ is in the first or second quadrant, $\tan{(a+b)}$ can be either positive or negtive. Using the addition formula for tangent I get
$$\tan{(a+b)}=\frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}=\frac{2+3}{1-2\cdot 3}=\frac{5}{-5}=-1.$$
This tells me that $\pi/2 < \alpha + \beta < \pi.$ So,
$$\alpha+\beta=\arctan{(-1)}+\pi k=-\frac{\pi}{4}+\pi k \ , \ \ k\in\mathbb{Z}.$$
$n=1$ takes me to the desired quadrant: $$\alpha + \beta =-\frac{\pi}{4}+\pi=\frac{3\pi}{4}.$$
Question: If $\tan{\beta}=-1/3,$ would the answer then have been $$\alpha + \beta= \frac{\pi}{4} \ \ ?$$
