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I have tried and tried but cannot for the life of me see how one equation follows onto the other... can anybody help??

$$\Omega(\theta)=-b \coth(\operatorname{arsinh}(e^{a\theta} \sinh c_0 ))$$

$$\implies \Omega(\theta)=\sqrt{c_1 e^{-2a\theta} + b^2}$$

note that: $c_0=-\operatorname{arcoth}\left(\frac{\Omega_0}{b}\right)$

Eichberger quotes (at the bottom of page 5) "By inserting (11) into (8), using identities of the hyperbolic transcendental functions and carefully observing $±$ signs, we obtain $Ω$ as a function of $θ$."

http://www.roulette.gmxhome.de/roulette%5B1%5D.pdf

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$\DeclareMathOperator{\arsinh}{arsinh}$I haven't looked at your PDF, but here are some observations:

Since $\cosh^{2} u - \sinh^{2} u = 1$ and $\cosh u > 0$ for all real $u$, we have $\cosh u = \sqrt{1 + \sinh^{2} u}$, and therefore $$ \coth(\arsinh x) = \frac{\cosh(\arsinh x)}{\sinh(\arsinh x)} = \frac{\sqrt{1 + x^{2}}}{x} = \sqrt{1 + \frac{1}{x^{2}}} $$ for all real $x \neq 0$. Since it appears you have $x = \exp(a\theta) \sinh(c_{0})$, substitution gives \begin{align*} \Omega(\theta) &= -b \coth\bigl[\arsinh\bigl(\exp(a\theta) \sinh(c_{0})\bigr)\bigr] \\ &= -b \sqrt{1 + \frac{1}{\exp(2a\theta) \sinh^{2}(c_{0})}} \\ &= - \sqrt{b^{2} + \frac{b^{2}}{\sinh^{2}(c_{0})}\exp(-2a\theta)}, \end{align*} which (up to a sign) has the general form you're seeking.

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