# $\frac{\pi}{2} =\tan^{-1}(\infty)$

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?

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Yes, you need to take limits, and you need to approach $\frac{\pi}{2}$ from the left (the two-sided limit isn't well-defined even in the extended reals). Alternately, a more abstract fix is to make the tangent take values in something other than the reals (for example the real projective line, which is where it naturally takes values). – Qiaochu Yuan Jun 17 '11 at 12:55

## 3 Answers

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.

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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.

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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 $\pi/2$.

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Thanks for the reply, your explanation of $\tan(\alpha_1 + \alpha_2)$ not existing was very helpful. :) – amy Jun 17 '11 at 14:59