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I'm working through James Stewart's Precalculus, and I have some confusion regarding this question: "Find the exact value of the trigonometric function at the given real number: $\cot \frac{25\pi}2$."

So, easy enough... $\frac{25\pi}2$ is simply a multiple of $\pi\over2$. $\cot$ is undefined at intervals of $n\pi$ where $n$ is any integer, and the value of $\cot$ at $\pi\over2$ is $0$, as evidenced by the graph below: Graph of Cotangent

Now comes the part, that befuddles me: working it out algebraically, I do the following.

$$\begin{align} \cot \frac{25\pi}2 &= \cot \frac\pi2 \\[4pt] &= \frac{1}{\tan(\pi/2)} \\[4pt] &= \frac{1}{\text{undefined}} \end{align}$$

The value of $\tan$ for $\pi\over2$ is, of course, undefined, as evidenced by the graph below: enter image description here

It is here that I get stuck:

How is that $\cot$ is defined for that value when $\tan$ itself is not?

I would imagine that a numerator divided by a denominator of an undefined value would be undefined? Could anyone explain how the value of 0 for cotangent would be worked out in terms of tangent. Is it incorrect to relate cotangent and tangent like this:

$$cot = \frac1{tan}$$

I hope that makes sense and is not too a silly a question.

Thanks in advance!

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    $\begingroup$ $\frac1{\tan\theta}$ may be undefined at $\frac\pi2$, but $\frac{\cos\theta}{\sin\theta}$ isn’t. If you don’t like that, then you’ll have to take $\lim_{\theta\to\frac\pi2}\frac1{\tan\theta}$. $\endgroup$ – amd Nov 19 '15 at 11:28
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    $\begingroup$ Speaking rather loosely, $\frac{1}{\tan{(\pi /2)}} = \frac{1}{\pm \infty} = \frac{1}{\mathtt{REALLY BIG}} = 0$. (But this is sloppy math. Go with amd's suggestion above.) $\endgroup$ – zahbaz Dec 1 '15 at 5:54
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The standard definition of $\cot x$ is $\frac{\cos x}{\sin x}$. This happens to coincide with $\frac{1}{\tan x}$ whenever $\cos x \neq 0$. Defining $\cot x$ as $\frac{1}{\tan x}$ without mentioning this is just lazy.

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$\tan({\pi\over 2})$ is undefined because $\cos({\pi\over 2}) = 0$

But $$\cot \left({\pi\over 2}\right) = {\cos({\pi\over 2})\over\sin({\pi\over 2})} = {0\over 1} = 0$$

Now $\cot(\theta) = (\tan(\theta))^{-1}$ because

$$\cot \left(\theta\right) = {\cos(\theta)\over\sin(\theta)} = \left({\sin(\theta)\over\cos(\theta)}\right)^{-1} = (\tan\left(\theta\right))^{-1}$$

it is not the other way around, just because $\tan^{-1}(\theta)$ works for most value does not make it the definition.

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  • $\begingroup$ $tan^{-1}x$ and $(\tan x)^{-1}$ are different functions. First one is $\arctan$ and other one is $\frac{1}{\tan x}$. Please don't use confusing notations. $\endgroup$ – Jaideep Khare Jul 5 '17 at 6:54
  • $\begingroup$ @JaideepKhare Changed, Thanks. But I believe that people can understand the difference here. $\endgroup$ – A---B Jul 5 '17 at 9:27
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If you take the left limit of $1/tan$ you'll get $0$. If you take the right limit of $1/tan$ you will also get $0$, since you are dividing by really big things $\infty$ and $-\infty$ respectively. Since both limits agree, the limit exists and it's $0$.

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$$ \cot \frac{25 \pi}{2}= \frac{\cos (25 \pi/2)}{\sin (25 \pi/2)}= \frac{0}{1}=0$$

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