# Equivalence of Definitions of Real Area Hyperbolic Secant

## Theorem

The following definitions of the concept of Real Area Hyperbolic Secant are equivalent:

### Definition 1

The inverse hyperbolic secant $\sech^{-1}: S \to \R$ is a real function defined on $S$ as:

$\forall x \in S: \map {\sech^{-1} } x := y \in \R_{\ge 0}: x = \map \sech y$

where $\map \sech y$ denotes the hyperbolic secant function.

### Definition 2

The inverse hyperbolic secant $\sech^{-1}: S \to \R$ is a real function defined on $S$ as:

$\forall x \in S: \map {\sech^{-1} } x := \map \ln {\dfrac {1 + \sqrt {1 - x^2} } x}$

where:

$\ln$ denotes the natural logarithm of a (strictly positive) real number.
$\sqrt {1 - x^2}$ denotes the positive square root of $1 - x^2$

## Proof

### Definition 1 implies Definition 2

 $\ds x$ $=$ $\ds \sech y$ $\ds \leadsto \ \$ $\ds \dfrac 1 x$ $=$ $\ds \cosh y$ Definition of Hyperbolic Secant $\ds \leadsto \ \$ $\ds y$ $=$ $\ds \map \ln {\dfrac 1 x + \sqrt {\paren {\dfrac 1 x}^2 - 1} }$ Definition of Real Inverse Hyperbolic Cosine $\ds$ $=$ $\ds \map \ln {\dfrac 1 x + \sqrt {\dfrac {1 - x^2} {x^2} } }$ $\ds$ $=$ $\ds \map \ln {\dfrac 1 x + \dfrac {\sqrt {1 - x^2} } x}$ as $x > 1$ $\ds$ $=$ $\ds \map \ln {\dfrac {1 + \sqrt {1 - x^2} } x}$

$\Box$

### Definition 2 implies Definition 1

 $\ds y$ $=$ $\ds \map \ln {\dfrac {1 + \sqrt {1 - x^2} } x}$ $\ds$ $=$ $\ds \map \ln {\dfrac 1 x + \sqrt {\dfrac {1 - x^2} {x^2} } }$ $\ds$ $=$ $\ds \map \ln {\dfrac 1 x + \sqrt {\paren {\dfrac 1 x}^2 - 1} }$ $\ds$ $=$ $\ds \arcosh \dfrac 1 x$ Definition of Real Area Hyperbolic Cosine $\ds \leadsto \ \$ $\ds \dfrac 1 x$ $=$ $\ds \cosh y$ Definition of Hyperbolic Cosine $\ds \leadsto \ \$ $\ds x$ $=$ $\ds \sech y$ Definition of Hyperbolic Secant

$\Box$

Therefore:

 $\text {(1)}: \quad$ $\ds x = \sech y$ $\implies$ $\ds y = \map \ln {\dfrac {1 + \sqrt {1 - x^2} } x}$ Definition 1 implies Definition 2 $\text {(2)}: \quad$ $\ds y = \map \ln {\dfrac {1 + \sqrt {1 - x^2} } x}$ $\implies$ $\ds x = \sech y$ Definition 2 implies Definition 1 $\ds \leadsto \ \$ $\ds x = \sech y$ $\iff$ $\ds y = \map \ln {\dfrac {1 + \sqrt {1 - x^2} } x}$

$\blacksquare$