Given an analytic result from solving an equation, we have the result:
$$\frac{1-\operatorname{cosh}(L)}{\operatorname{sinh}(L)} \operatorname{cosh}(x) + \operatorname{sinh}(x).$$
This can be shown to be equivalent (e.g. with Mathematica) to the expression:
$$\frac{\operatorname{sinh}\left(x-\frac{L}{2}\right)}{\operatorname{cosh}\left(\frac{L}{2}\right)}.$$
Despite this just being a simple manipulation of hyperbolic identities, I cannot reach the latter expression from the first. Any advice/approach gratefully appreciated.
Mathematica finds an intermediate stage to be:
$$\operatorname{sinh}\left(x\right) - \operatorname{cosh}\left(x\right)\operatorname{tanh}\left(\frac{L}{2}\right).$$
Again, I can't see how to obtain this from the first result. Maybe I need some more coffee... I have tried using the double angle formulas for sinh and cosh, inappropriately squaring etc - I am not sure how to proceed from here! I am completely happy with the equivalence, but just would like to derive it on paper!