Using the result, $\tan^2{\alpha} - A \tan{\alpha} + 1 = 0~$, where A is a constant, prove that the two solutions to this equation (such that $0 \leq \alpha \leq \frac{\pi}{2}$) are complementary (i.e. $\alpha_1 + \alpha_2=\large \frac{\pi}{2}~$)
To solve this equation, suppose the two roots are $\alpha_1$ and $\alpha_2$. We can take the product of roots $\Pi\tan{\alpha}: \tan{\alpha_1}\tan{\alpha_2}=1~$, but noting that $\tan(\alpha_1 + \alpha_2) = \large \frac{\tan{\alpha_1}+\tan{\alpha_2}}{1 - \tan{\alpha_1}\tan{\alpha_2}}~$ from the $\tan$ expansion.
Substituting $\tan{\alpha_1}\tan{\alpha_2}=1~$ into that expansion:
$\tan(\alpha_1 + \alpha_2) = \frac{\tan{\alpha_1}+\tan{\alpha_2}}{1 - 1}=\frac{\tan{\alpha_1}+\tan{\alpha_2}}{0}=\infty~$
Therefore, following on from this, $\alpha_1 + \alpha_2 = \tan^{-1}(\infty) = \large \frac{\pi}{2}$
However, is this proof flawed, especially in equating $\large \frac{\tan{\alpha_1}+\tan{\alpha_2}}{0}$ with $\infty~$, and $\tan^{-1}(\infty)$ with $\large \frac{\pi}{2}~$? Would limits be required for a more proper treatment?