# What is $\cot(\pi/2)$?

Base on the unit circle, I know

\begin{align} &\cot\left(\frac{\pi}{2}\right) \\ =&\frac{0}{1}\\ =&0 \end{align}

But it is also

\begin{align} &\cot\left(\frac{\pi}{2}\right) \\ =&\frac{1}{\tan\left(\frac{\pi}{2}\right)}\\ =&\frac{1}{\frac{1}{0}}\\ =&undefined \end{align}

$6.12303177 × 10^{-17}$

I am really confused now. Although I know it is $0$, I don't see why the other ones are wrong.

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What do you mean? The other ones are wrong because they disagree with the one that's right. –  Gerry Myerson Oct 14 '12 at 22:44
@GerryMyerson - I mean which step is wrong. –  Derek 朕會功夫 Oct 14 '12 at 22:46
Well, Google doesn't give any steps, so I don't know what you mean by "which step is wrong". For the other one, you could say that $1/0$ isn't undefined, it's infinite, and $1/{\infty}=0$. –  Gerry Myerson Oct 14 '12 at 22:49
@GerryMyerson - Why $1/∞$ is $0$? –  Derek 朕會功夫 Oct 14 '12 at 22:51
Is it because of $lim_{(x\to∞)} \frac{1}{x} = 0$? –  Derek 朕會功夫 Oct 14 '12 at 22:58

$$\cot x = \frac{1}{\tan x}$$ only when $\tan x \neq 0$ (i.e. $x \neq n\pi$ for any $n\in \mathbb {Z}$).

However, $\cot x$ is actually defined as

$$\cot x := \frac{\cos x}{\sin x}$$ so $\cot \left(\frac{\pi}{2}\right)=0$ is the correct answer.

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To add to Argon's answer, the Google answer is due to numerical round-off error. –  Ganesh Oct 14 '12 at 22:44

$\tan \frac π2$ is actually infinity.

So if we divide 1 by $\tan \frac π2$ we will have $\frac 1{\infty}$ which leads to 0.

$$\cot \frac π2 = \frac 1{\tan \frac π2}=0$$

Furthermore:

$$\cot x=\frac 1{\tan x}=\frac{\cos x}{\sin x}$$

So:

$$\cot \frac π2=\frac{\cos \frac π2}{\sin \frac π2} = \frac 01 = 0$$

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