Inverse Hyperbolic secant function

The inverse form of the hyperbolic secant function is called the inverse hyperbolic secant function.


$\large {\mathop{\rm sech}\nolimits}^{-1}{x} \,=\, \log_{e}{\Bigg(\dfrac{1+\sqrt{1-x^2}}{x}\Bigg)}$

The hyperbolic secant function is defined as the ratio of $2$ to sum of positive and negative natural exponential functions. Hence, the inverse of hyperbolic secant function should be defined in the form of natural logarithmic function. It can be derived in mathematics on the basis of the definition of the hyperbolic secant function.


Assume, $x$ and $y$ are two literal numbers and also assume the value of $x$ is equal to the hyperbolic secant of $y$.

$x = {\mathop{\rm sech}\nolimits}{y}$

Therefore, the value of $y$ should be the inverse hyperbolic secant of $x$. It is written as ${\mathop{\rm sech}\nolimits}^{-1}{x}$ or ${\mathop{\rm arsech}\nolimits}{x}$

$y = {\mathop{\rm sech}\nolimits}^{-1}{x} \,\,\,$ (or) $\,\,\, {\mathop{\rm arsech}\nolimits}{x}$

The mathematical relation between hyperbolic and inverse hyperbolic secant functions can be written as follows.

$x = {\mathop{\rm sech}\nolimits}{y} \,\,\Leftrightarrow \,\, y = {\mathop{\rm sech}\nolimits}^{-1}{x} \,\,\,$ (or) $\,\,\, {\mathop{\rm arsech}\nolimits}{x}$

Transforming Hyperbolic function in exponential form

According to the definition of the hyperbolic secant function

$x \,=\, \dfrac{2}{e^y+e^{-y}}$

Simplification of the equation

$\implies$ $x(e^y+e^{-y}) \,=\, 2$

$\implies$ $x\Big(e^y+\dfrac{1}{e^y}\Big) \,=\, 2$

$\implies$ $x\Big(\dfrac{e^{2y}+1}{e^y}\Big) \,=\, 2$

$\implies$ $x(e^{2y}+1) \,=\, 2e^y$

$\implies$ $xe^{2y}+x \,=\, 2e^y$

$\implies$ $xe^{2y}+x-2e^y \,=\, 0$

$\implies$ $xe^{2y}-2e^y+x \,=\, 0$

It is a quadratic equation and solve this equation to obtain the expression of the inverse hyperbolic secant function in the form of a logarithmic function.

Solving the quadratic equation

$\implies$ $x{(e^y)}^2-2e^y+x \,=\, 0$

Take $m = e^y$ for simplifying the equation easily.

$\implies$ $xm^2-2m+x \,=\, 0$

Use quadratic formula method and express the roots of this quadratic equation.

$m \,=\, \dfrac{-(-2) \pm \sqrt{{(-2)}^2-4 \times x \times x}}{2 \times x}$

$\implies$ $m \,=\, \dfrac{2 \pm \sqrt{4-4x^2}}{2x}$

$\implies$ $m \,=\, \dfrac{2 \pm \sqrt{4(1-x^2)}}{2x}$

$\implies$ $m \,=\, \dfrac{2 \pm 2\sqrt{1-x^2}}{2x}$

$\implies$ $m \,=\, \dfrac{2(1 \pm \sqrt{1-x^2})}{2x}$

$\implies$ $\require{cancel} m \,=\, \dfrac{\cancel{2}(1 \pm \sqrt{1-x^2})}{\cancel{2}x}$

$\implies$ $m \,=\, \dfrac{1 \pm \sqrt{1-x^2}}{x}$

Actually, $m = e^y$. So, replace $m$ by its actual value.

$\,\,\, \therefore \,\,\,\,\,\,$ $e^y \,=\, \dfrac{1 \pm \sqrt{1-x^2}}{x}$

Therefore, $e^y \,=\, \dfrac{1+\sqrt{1-x^2}}{x}$ and $e^y \,=\, \dfrac{1-\sqrt{1-x^2}}{x}$.

Verifying the solutions

The value of $e$ is a positive irrational number. So, the value of $e^y$ is always a positive number and it is possible only by the expression $\dfrac{1+\sqrt{1-x^2}}{x}$. The remaining solution $\dfrac{1-\sqrt{1-x^2}}{x}$ always gives negative values due to the negative sign in the numerator. Therefore, $\dfrac{1+\sqrt{1-x^2}}{x}$ is the true solution of the $e^y$ mathematically.

$\,\,\, \therefore \,\,\,\,\,\,$ $e^y \,=\, \dfrac{1+\sqrt{1-x^2}}{x}$

Obtaining the inverse form of hyperbolic function

Take natural logarithm both sides to simplify the mathematical equation further.

$\log_{e}{e^y} \,=\, \log_{e}{\Bigg(\dfrac{1+\sqrt{1-x^2}}{x}\Bigg)}$

According to the power rule of the logarithms,

$\implies$ $y \times \log_{e}{e} \,=\, \log_{e}{\Bigg(\dfrac{1+\sqrt{1-x^2}}{x}\Bigg)}$

$\implies$ $y \times 1 \,=\, \log_{e}{\Bigg(\dfrac{1+\sqrt{1-x^2}}{x}\Bigg)}$

$\implies$ $y \,=\, \log_{e}{\Bigg(\dfrac{1+\sqrt{1-x^2}}{x}\Bigg)}$

The value of $y$ in terms of $x$ is equal to ${\mathop{\rm sech}\nolimits}^{-1}{x}$.

$\,\,\, \therefore \,\,\,\,\,\,$ ${\mathop{\rm sech}\nolimits}^{-1}{x} \,=\, \log_{e}{\Bigg(\dfrac{1+\sqrt{1-x^2}}{x}\Bigg)}$

Follow us
Email subscription