$\frac{\pi}{2}  =\tan^{-1}(\infty)$ 
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?
 A: We need to make the question more precise.  After all, for any real number $u$, there are infinitely many $\alpha$ such that $\tan \alpha=u$.  So we will restate the problem.
Suppose that $A>0$ and $A^2-4 \ge 0$.  Then the equation 
$$x^2 -Ax+1=0$$ 
has two (possibly equal) positive solutions, say $x_1$ and $x_2$.
Let $\alpha_i$ be the angle (number) in the interval $(0,\pi/2)$ such that $\tan\alpha_i=x_i$.  Show that $\alpha_1 +\alpha_2=\pi/2$.
The addition formula for $\tan$ seems to me not the best way to handle the problem, even though, as Gerry Myerson points out, there is a precise way of interpreting the situation when the denominator is $0$.
It seems to me simpler to note that if $\alpha_i$ are angles in the interval $(0,\pi/2)$ we have
$$\alpha_1+\alpha_2=\frac{\pi}{2} \qquad \text{iff}\qquad \tan\alpha_2=\frac{1}{\tan\alpha_1}$$
This follows from primitive properties of right triangles.  
So the only thing to verify is that the product of the roots of $x^2-Ax+1=0$ is $1$, and this is clear from the shape of the equation.  
Added comment: If one takes great care, or if one has very good intuition, it is possible to handle "$\infty$" without making errors.  Euler (mostly) did it, but we are not all Euler. Improper handling of "$\infty$" is an all too frequent source of student mistakes.  So it is best to do "defensive thinking," and avoid trying to handle "$\infty$" as if it were a number. After all, it isn't. 
A: Put $z = \tan(\alpha)$.   Then your equation is 
$$z^2 - Az + 1 = 0.$$
Real roots exist provided that $A^2 - 4 \ge 0$  In this case, the roots are reciprocals.  Your result then follows right away, since tangents of complementary angles are necessarily reciprocals.
A: I don't think there's any need for limits, nor for projective planes. The formula doesn't show $\tan(\alpha_1+\alpha_2)$ is infinite, it shows that it doesn't exist, and the argument for which the tangent doesn't exist is $\pi/2$. 
