$\frac{1}{\frac{1}{\tan{0}}}$. Is it zero or undefined?

Question

I had a quiz in algebra 2, and it was all about trig. There was this question that was asking to find the trig value of a fraction involving cotangent, and I ended up having $\frac{1}{\frac{1}{\tan{0}}}$ after simplifying. Since $\tan(0)=0$, I answered the answer for the question was undefined because it is not allowed to have a value of $0$ in the denominator. However, the answer was $0$, changing $\tan(0)$ to $0$ after simplifying to $\tan(0)$. I think we can't keep on calculating once the denominator becomes zero. Here's what I have done. The original fraction was $(\cot(\frac{5\pi}{2}))^2$ and I changed its form to 1/(tan(5pi/2))^2 which becomes 1/(cot(0))^2 and resulting in 1/1/(-tan (0))^2 And, I answered the answer for the question was undefined because it is not allowed to have a value of $0$ in the denominator. What's wrong in my process?

Answer 1

Piecing together some comments, it appears the OP did something along the lines of the following analysis:

$$\cot(5\pi/2)={1\over\tan(5\pi/2)}={1\over\tan\left({5\pi\over2}+0\right)}={1\over-\cot0}={1\over-{1\over\tan0}}$$

and then concluded (quite correctly!) that the final expression is undefined because you are not ever allowed to divide by $0$.

The resolution here is to keep in mind that the trig identity

$$\cot x={1\over\tan x}$$

technically holds only when $x$ in the domain of both the tangent and the cotangent function. So in fact the very first equality in the OP's (ostensible) analysis is technically invalid, since $5\pi/2$ is not in the domain of the tangent function. In other words, the appropriate thing to have done, as others have pointed out, is to use the definition of the cotangent,

$$\cot(5\pi/2)={\cos(5\pi/2)\over\sin(5\pi/2)}={0\over1}=0$$

which is simpler anyway, avoiding as it does the trig identity $\tan\left({5\pi\over2}+\theta\right)=-\cot\theta$, which I think underlies one step in what the OP did. (I'm impressed the OP got the minus sign, because I had to think really hard to see it.)

Note, I've said "technically" here a couple of times. That's because you can often reason informally that $0={1\over\pm\infty}$ and/or $\pm\infty={1\over0}$, so that

$${1\over{-{1\over\tan0}}}={1\over-{1\over0}}={1\over\mp\infty}=0$$

(and it's even possible to formalize this reasoning). Alternatively, if the OP had simply kept simplifying, they would would have arrived at

$$\cot(5\pi/2)={1\over\tan(5\pi/2)}={1\over\tan\left({5\pi\over2}+0\right)}={1\over-\cot0}={1\over-{1\over\tan0}}=-\tan0=0$$

Answer 2

The problem that you had, I think, is that you tried to perform the computation as $$ \cot\left( \frac{5\pi}{2} \right)^2 = \frac{1}{\tan\left( \frac{5\pi}{2} \right)^2}. $$ In general, if $x$ is in the domain of both $\cot$ and $\frac{1}{\tan}$, then this will work and will be a perfectly reasonable argument. That is, for such $x$, we definitely have $$ \cot(x)^2 = \frac{1}{\tan(x)^2}. $$ However, $\frac{(2k+1)\pi}{2}$ is not in the domain of $\frac{1}{\tan}$ for any integer $k$, which means that this computation doesn't really work.

You might have better luck with your computation if you do it as follows: $$ \cot\left( \frac{5\pi}{2} \right)^2 = \frac{\cos\left( \frac{5\pi}{2} \right)^2}{\sin\left( \frac{5\pi}{2} \right)^2} = \frac{0}{1} = 0. $$