# Definition:Inverse Secant/Arcsecant

## Definition

### Real Numbers

Arcsecant Function

From Shape of Secant Function, we have that $\sec x$ is continuous and strictly increasing on the intervals $\left[{0 \,.\,.\, \dfrac \pi 2}\right)$ and $\left({\dfrac \pi 2 \,.\,.\, \pi}\right]$.

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 $g: \left[{0 \,.\,.\, \dfrac \pi 2}\right) \to \left[{1 \,.\,.\, \infty}\right)$ be the restriction of $\sec x$ to $\left[{0 \,.\,.\, \dfrac \pi 2}\right)$.

Let $h: \left({\dfrac \pi 2 \,.\,.\, \pi}\right] \to \left({-\infty \,.\,.\, -1}\right]$ be the restriction of $\sec x$ to $\left({\dfrac \pi 2 \,.\,.\, \pi}\right]$.

Let $f: \left[{0 \,.\,.\, \pi}\right] \setminus \dfrac \pi 2 \to \R \setminus \left({-1 \,.\,.\, 1}\right)$:

$f\left({x}\right) = \begin{cases} g\left({x}\right) & : 0 \le x < \dfrac \pi 2 \\ h\left({x}\right) & : \dfrac \pi 2 < x \le \pi \end{cases}$

From Inverse of Strictly Monotone Function, $g \left({x}\right)$ admits an inverse function, which will be continuous and strictly increasing on $\left[{1 \,.\,.\, \infty}\right)$.

From Inverse of Strictly Monotone Function, $h \left({x}\right)$ admits an inverse function, which will be continuous and strictly increasing on $\left({-\infty \,.\,.\, -1}\right]$.

As both the domain and range of $g$ and $h$ are disjoint, it follows that:

$f^{-1}\left({x}\right) = \begin{cases} g^{-1}\left({x}\right) & : x \ge 1 \\ h^{-1}\left({x}\right) & : x \le -1 \end{cases}$

This function $f^{-1} \left({x}\right)$ is called arcsecant of $x$ and is written $\operatorname{arcsec} x$.

Thus:

• The domain of $\operatorname{arcsec} x$ is $\R \setminus \left({-1 \,.\,.\, 1}\right)$
• The image of $\operatorname{arcsec} x$ is $\left[{0 \,.\,.\, \pi}\right] \setminus \dfrac \pi 2$.

### Complex Plane

The principal branch of the complex inverse secant function is defined as:

$\forall z \in \C_{\ne 0}: \map \arcsec z := \dfrac 1 i \, \map \Ln {\dfrac {1 + \sqrt {1 - z^2} } z}$

where:

$\Ln$ denotes the principal branch of the complex natural logarithm
$\sqrt {1 - z^2}$ denotes the principal square root of $1 - z^2$.