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During my recent study, I found an Identity which is of the form $$ \tanh(N\textrm{acosh}\;a) = \left\vert \frac{g^{2N}-1}{g^{2N}+1}\right\vert $$ where $a\geq1$ and $g>0$ satisfy $a=\frac{g^2+1}{2g}$. I do not know how to prove it, but my motivation was presented in my previous question. In the previous question my attempts had been shown.

How about the identity? Could you show me how to prove it?


Im Sorry, but I know how to prove it now. The main idea lies in that, if $g=a+\sqrt{a^2-1}$, then $$ \tanh(N\ln g) = \tanh(N\ln(a+\sqrt{a^2-1})) = \tanh(N\textrm{acosh}\;a) $$ and by the substitution $x=N\ln g$ $$ \tanh(N\ln g) = \tanh x = \frac{e^x-e^{-x}}{e^x+e^{-x}} = \frac{g^N-g^{-N}}{g^N+g^{-N}}=\frac{g^{2N}-1}{g^{2N}+1}. $$

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  • $\begingroup$ If you wish, you can answer your own question and accept your own answer. $\endgroup$ – Vincenzo Oliva Jan 26 '16 at 15:09

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