What is $\cot(\pi/2)$? Base on the unit circle, I know 

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

But it is also

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

And Google gives me this answer:

　$6.12303177 × 10^{-17}$

I am really confused now. Although I know it is $0$, I don't see why the other ones are wrong.
$$\cot x = \frac{1}{\tan x}$$ 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.
$\tan \frac π2$ 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: 

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