When working on the following equation for my pre-calculus class a couple nights ago

Verify that: $$ \tan \left(\frac{\pi}{2} - x\right) = \cot x $$

I decided to answer the question by using the fact that $$ \tan \left(\frac{\pi}{2} - x\right) = \frac{\tan \frac{\pi}{2} - \tan x}{1 + \tan \frac{\pi}{2} \tan x}. $$

I realized that using this formula, I would not help me answer the question for my math class, because, as we learned in class, $\tan \frac{\pi}{2}$ is undefined. However, I (knowingly erroneously) proceeded with this method, plugging in $\pm \infty$ for $\tan \frac{\pi}{2}$ yielding $$ \frac{\pm \infty - \tan x}{1 + \pm \infty \tan x} $$

which can be simplified down to $$ \frac{1 - 0}{0 + \tan x} = \frac{1}{\tan x} = \cot x $$

by dividing the numerator and the denominator by $\pm \infty$.

In an attempt to justify the fact that $\frac{\infty}{\infty}$ should be 1 and $\frac{1}{\infty}$ and $\frac{\tan x}{\infty}$ should be 0, I turned to limits. Whilst not knowing too much calculus, I learned last year in Algebra II that $\infty$ can be somewhat approximated with limits. As such, I defined a new constant $\beta$ as $$ \beta = \lim_{x \rightarrow \infty} x $$

and $\alpha$ as $\beta$'s reciprocal. From my understanding of limits, that just means $\beta$ is essentially some very large number, so I should still be able to do arithmetic with it ($\alpha \beta = 1$, $\beta - \beta = 0$, etc.).

So then my equation becomes: $$ \frac{\beta - \tan x}{1 + \beta \tan x} = \frac{\beta - \tan x}{1 + \beta \tan x} \cdot \frac{\alpha}{\alpha} = \frac{1 - \alpha \tan x}{\alpha + \tan x} $$

which is essentially $\cot x$ at the end, because $\alpha \approx 0$.

I have been constantly told not to use infinities in equations since elementary school, yet here it seems to yield the correct answer. Why is that, and when can these infinities be used in equations? If I got this wrong, where did I err?

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    $\begingroup$ You can avoid infinities by expressing $\tan\left(\frac{\pi}{2} - x\right)$ as $\frac{\sin\left(\frac{\pi}{2} - x\right)}{\cos\left(\frac{\pi}{2} - x\right)}$, then simplifying using the complementary angle identities. $\endgroup$ Commented Oct 31, 2021 at 1:27
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    $\begingroup$ @N.F.Taussig Thanks for the advice. I can't believe I didn't realize that! I ended up using the difference formulas for sine and cosine to solve the problem. $\endgroup$ Commented Oct 31, 2021 at 1:30

2 Answers 2


Unfortunately, your justification for writing $\frac{\infty}{\infty}=1$ is not enough. There are several instances of ratios of functions $\frac{f(x)}{g(x)}$ whose numerator and denominator both approach $\infty$ as $x$ approaches some number $a$, but $\lim_{x\to a}\frac{f(x)}{g(x)}$ exists as a finite number. One example is obtained by letting $f(x)=\frac{1}{|x|}$, $g(x)=\frac{1}{x^2}$, and $a=0$. If you compute the values of $f$ and $g$ near $a$, you'll see that $f(x)$ and $g(x)$ both explode to $\infty$ as $x$ gets closer and closer to $0$, but


and $|x|$ gets closer and closer to $0$ as $x$ approaches $0$, "giving" $\frac{\infty}{\infty}=0$. More generally, the expression $\frac{\infty}{\infty}$ can be made equal to any number you like by using suitable limits, and is thus usually left undefined.

That said, we can make rigorous your "division by $\infty$" at the beginning, and avoid the use of "infinite" numbers like $\beta$ and infinitesimals like $\alpha$, by using a limit. Instead of jumping from

$$\tan\left(\frac{\pi}{2}-x\right)\text{ to }\frac{\tan(\pi/2)-\tan(x)}{1+\tan(\pi/2)\tan(x)}$$

and suffering problems with $\infty$, start with the fact that $\tan(\pi/2-x)$ can be approximated by $\tan(h-x)$ when $h$ is near $\pi/2$, and that this approximation gets better and better as $h$ gets closer and closer to $\pi/2$. More precisely, we start with

$$\tan\left(\frac{\pi}{2}-x\right)=\lim_{h\to \frac{\pi}{2}}\tan(h-x)\text{ (assuming that }x\text{ is not an integer muliple of }\pi\text{)}$$

and then apply the tangent sum identity to get

$$\tan(h-x)=\frac{\tan(h)-\tan(x)}{1+\tan(h)\tan(x)}\text{ for every }h\text{ sufficiently close to }\frac{\pi}{2}$$

Why the words "sufficiently close"? Well, for any given $x$, $\tan(h-x)$ is undefined if $h-x$ is an odd integer multiple of $\pi/2$, that is, $h-x=(2n+1)\frac{\pi}{2}$ for some integer $n$. Since we want to study the behavior of $\tan(h-x)$ when $h$ is close to $\pi/2$, but also want everything to be well-defined, we need to constrain $h$.

With this settled, if $h$ is close enough to $\frac{\pi}{2}$ to never be an integer multiple of $\pi$, then $\tan(h)$ will never be zero, so we can divide the numerator and denominator by the "infinite" number $\tan(h)$ (infinite because $\tan(h)$ grows to $\pm\infty$ as $h$ gets closer and closer to $\pi/2$) to get

$$\frac{\tan(h)-\tan(x)}{1+\tan(h)\tan(x)}=\frac{1-\frac{\tan(x)}{\tan(h)}}{\frac{1}{\tan(h)}+\tan(x)}\text{ for every }h\text{ sufficiently close to }\frac{\pi}{2}$$

As we previously mentioned, $\tan(h)$ explodes to $\pm\infty$ as $h$ approaches $\pi/2$. Bringing in the problematic infinities for suggestive notation, this gives

$$\frac{1-\frac{\tan(x)}{\tan(h)}}{\frac{1}{\tan(h)}+\tan(x)}\to\frac{1-\frac{\tan(x)}{\pm\infty}}{\frac{1}{\pm\infty}+\tan(x)}=\frac{1-0}{0+\tan(x)}=\cot(x)\text{ as }h\text{ approaches }\frac{\pi}{2}$$

More precisely, what we've shown is that the limit of our expression as $h$ approaches $\pi/2$ is $\cot(x)$:


We can then conclude that $\tan(\pi/2-x)=\cot(x)$ because the limit of a function is necessarily unique.

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    $\begingroup$ It may be helpful to point out that $\tan\left(\frac{\pi}{2}-x\right)=\lim_{h\to \frac{\pi}{2}}\tan(h-x)$ because the function is continuous under your specified condition. $\endgroup$
    – rb612
    Commented Oct 31, 2021 at 2:49
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    $\begingroup$ @rb612 I did, didn't I? $\endgroup$ Commented Oct 31, 2021 at 2:52
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    $\begingroup$ Your logic is sound, but I think it's good to make note that $f(a)$ being defined is not necessarily sufficient for $f(a)=\lim_{x\to a} f(x)$, for someone in precalculus. $\endgroup$
    – rb612
    Commented Oct 31, 2021 at 2:59
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    $\begingroup$ @rb612 ah, I see what you mean. I'll edit my post to clarify this. $\endgroup$ Commented Oct 31, 2021 at 3:39

$$ \begin{aligned} \tan \left(\frac{\pi}{2}-x\right) &=\frac{\sin \left(\frac{\pi}{2}-x\right)}{\cos \left(\frac{\pi}{2}-x\right)} \\ &=\frac{\cos x}{\sin x} \\ &=\cot x \end{aligned} $$


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