\begin{definition}[Definition:Inverse Cosecant/Real/Arccosecant]
{{:Graph of Arccosecant Function}}
From Shape of Cosecant Function, we have that $\csc x$ is continuous and strictly decreasing on the intervals $\hointr {-\dfrac \pi 2} 0$ and $\hointl 0 {\dfrac \pi 2}$.
From the same source, we also have that:
:$\csc x \to + \infty$ as $x \to 0^+$
:$\csc x \to - \infty$ as $x \to 0^-$
Let $g: \hointr {-\dfrac \pi 2} 0 \to \hointl {-\infty} {-1}$ be the restriction of $\csc x$ to $\hointr {-\dfrac \pi 2} 0$.
Let $h: \hointl 0 {\dfrac \pi 2} \to \hointr 1 \infty$ be the restriction of $\csc x$ to $\hointl 0 {\dfrac \pi 2}$.
Let $f: \closedint {-\dfrac \pi 2} {\dfrac \pi 2} \setminus \set 0 \to \R \setminus \openint {-1} 1$:
:$\map f x = \begin{cases}
\map g x & : -\dfrac \pi 2 \le x < 0 \\
\map h x & : 0 < x \le \dfrac \pi 2
\end{cases}$
From Inverse of Strictly Monotone Function, $\map g x$ admits an inverse function, which will be continuous and strictly decreasing on $\hointl {-\infty} {-1}$.
From Inverse of Strictly Monotone Function, $\map h x$ admits an inverse function, which will be continuous and strictly decreasing on $\hointr 1 \infty$.
As both the domain and range of $g$ and $h$ are disjoint, it follows that:
:$\map {f^{-1} } x = \begin{cases}
\map {g^{-1} } x & : x \le -1 \\
\map {h^{-1} } x & : x \ge 1
\end{cases}$
This function $\map {f^{-1} } x$ is called '''arccosecant''' of $x$ and is written $\arccsc x$.
Thus:
:The domain of $\arccsc x$ is $\R \setminus \openint {-1} 1$
:The image of $\arccsc x$ is $\closedint {-\dfrac \pi 2} {\dfrac \pi 2} \setminus \set 0$.
\end{definition}