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

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

The hyperbolic cotangent function is defined in mathematics as the ratio of summation to subtraction of negative and positive natural exponential functions. The inverse form of the hyperbolic cotangent function is in the logarithmic function form. The inverse hyperbolic function can be derived mathematically from the hyperbolic cotangent function.

Take two literals $x$ and $y$ and the value of $x$ is equal to the hyperbolic cotangent of $y$.

$x = \coth{y}$

So, the value of $y$ should be the inverse hyperbolic cotangent of $x$.

$y = \coth^{-1}{x}$

The relation between the hyperbolic and inverse hyperbolic cotangent functions can be written in mathematical form as follows.

$x = \coth{y} \,\,\Leftrightarrow \,\, y = \coth^{-1}{x}$

01

Write the hyperbolic cotangent of $y$ in the form of combination of natural exponential functions.

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

02

Now, simplify the exponential algebraic equation and express the equation in terms of $x$ purely for evaluating the value of $y$.

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

$\implies$ $xe^y-xe^{-y} = e^y+e^{-y}$

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

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

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

$\implies$ $(1-x)e^y+(1+x)e^{-y} = 0$

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

$\implies$ $(1-x)e^y = -\dfrac{1+x}{e^y}$

$\implies$ $(1-x)e^y \times e^y = -(1+x)$

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

$\implies$ $e^{2y} = -\dfrac{1+x}{1-x}$

$\implies$ $e^{2y} = \dfrac{1+x}{x-1}$

$\implies$ $e^{2y} = \dfrac{x+1}{x-1}$

03

Eliminate the $y$ from the natural exponential function and it can be done by applying natural logarithmic system to both sides of the equation.

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

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

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

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

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

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

Latest Math Topics

Jan 06, 2023

Jan 03, 2023

Jan 01, 2023

Dec 26, 2022

Dec 08, 2022

Latest Math Problems

Nov 25, 2022

Nov 02, 2022

Oct 26, 2022

Oct 24, 2022

Sep 30, 2022

A best free mathematics education website for students, teachers and researchers.

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Learn how to solve the maths problems in different methods with understandable steps.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved