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