I’m trying to show that the involute of a catenary, calculated to be
$$\gamma(t)=(t-\tanh(t),\cosh(t)-\sinh(t)\tanh(t))$$ is a reparametrization of the tractrix
$$\beta(s)=\left(\cos(s)+\log\left(\tan\left(\frac{s}{2}\right)\right),\:\sin(s)\right).$$
I was suggested to go with $$\frac{1}{\cosh(t)}=\sin(s),$$
and that resolved the second component, but I’m absolutely stuck on the first part.
$$\text{If we set } \ T:=\tan(s/2),$$
$$\cos(s)=\frac{1-T^2}{1+T^2} \ \text{and} \ \sin(s)=\frac{2T}{1+T^2}\tag{1}$$
(1) are the classical "double angle formulas" ; you will find many references on them ; they are commonly used to convert trigonometrical espressions into polynomial expressions like in this answer.
Your curve description :
$$\left(\cos s +\ \ln (\tan \frac{s}{2}),\sin s \right)$$
becomes :
$$\left(\frac{1-T^2}{1+T^2}+\ln(T),\frac{2T}{1+T^2}\right)\tag{2}$$
If we set now $u:=\ln(T) \ \iff \ T=e^{u}$, (2) can be transformed into :
$$\left(\frac{1-e^{2u}}{1+e^{2u}}+u,\frac{2e^{u}}{1+e^{2u}}\right)\tag{3}$$
which (by dividing the numerator and the denominator of the fractions by $e^{u}$, is the same as
$$\left(\frac {e^{-u}-e^{u}}{e^{-u}+e^{u}}+u,\frac{2}{e^{-u}+e^{u}}\right)\tag{4}$$
Can you take it from here ?
Remark : of course, variable $T$ in (1) is not variable $t$ in your initial expression.
