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} $
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.
$\endgroup$ 82 Answers
$\begingroup$$$\cot x = \frac{1}{\tan x}$$ 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.
$\endgroup$ 3 $\begingroup$$\tan \frac π2$ is actually infinity.
So if we divide 1 by $\tan \frac π2$ we will have $\frac 1{\infty}$ which leads to 0.
$$\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$$
$\endgroup$ 1
