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 $θ$."