# hyperbolic function

• Jan 23rd 2008, 02:49 PM
nmanik90
hyperbolic function
how do you prove that

tanh^-1 [(x^2-a^2)/(x^2-a^2)] = ln (x/a) ?

• Jan 23rd 2008, 03:03 PM
Peritus
$\begin{gathered}
a\tanh \left( {\frac{{x^2 - a^2 }}
{{x^2 + a^2 }}} \right) = \ln \left( {\frac{x}
{a}} \right) \hfill \\
\hfill \\
\leftrightarrow \frac{{x^2 - a^2 }}
{{x^2 + a^2 }} = \tanh \left[ {\ln \left( {\frac{x}
{a}} \right)} \right] \hfill \\
\end{gathered}$

Now recall the following basic identity:

$\tanh x = \frac{{\sinh (x)}}
{{\cosh (x)}} = \frac{{\frac{{e^x - e^{ - x} }}
{2}}}
{{\frac{{e^x + e^{ - x} }}
{2}}} = \frac{{e^x - e^{ - x} }}
{{e^x + e^{ - x} }}$

thus:

$
\leftrightarrow \frac{{x^2 - a^2 }}
{{x^2 + a^2 }} = \frac{{e^{\ln \left( {\frac{x}
{a}} \right)} - e^{ - \ln \left( {\frac{x}
{a}} \right)} }}
{{e^{\ln \left( {\frac{x}
{a}} \right)} + e^{ - \ln \left( {\frac{x}
{a}} \right)} }} = \frac{{\frac{x}
{a} - \frac{a}
{x}}}
{{\frac{x}
{a} + \frac{a}
{x}}} = \frac{{x^2 - a^2 }}
{{x^2 + a^2 }}

$