I would like to prove that (1) $$\begin{equation} \tan\left(\frac{\theta}{2}\right)=\tan\left(\frac{\nu}{2}\right)\tan\left(\frac{\pi/2-\epsilon}{2}\right) \end{equation}$$
can transformed to (2) $$x=y+z,$$ where (3) \begin{align} x&=&\mathrm{arctanh}\left(cos(\theta)\right)\\y&=&\mathrm{arctanh}\left(cos(\nu)\right)\\z&=&\mathrm{arctanh}\left(\sin\left(\epsilon\right)\right) \end{align}
By solving for $\theta$ in 1 and 2, we see that these are indeed equal:
For the record, incorrect identity
Initially the question was wrongly stated, and the comments below pertain to this: I would like to prove that (1) $$\begin{equation} \tan\left(\frac{\theta}{2}\right)=\tan\left(\frac{\nu}{2}\right)\tan\left(\frac{\pi/2-\epsilon}{2}\right) \end{equation}$$
can transformed to (2) $$x=y+z,$$ where (3) \begin{align} x&=&\mathrm{arctanh}\left(\theta\right)\\y&=&\mathrm{arctanh}\left(\nu\right)\\z&=&\mathrm{arctanh}\left(\sin\left(\epsilon\right)\right) \end{align}
My attempt on incorrect identity:
Taking the tanh of (2) on both sides and using $\begin{align} \tanh(x + y) &= \frac{\tanh x +\tanh y}{1+ \tanh x \tanh y } \end{align}$ results in (2a) $$\boxed{ \theta = \frac{\nu+\sin\left(\epsilon\right)}{1+\nu \sin\left(\epsilon\right)} }$$
On the other hand, using
\begin{align} \tan \frac{\theta}{2} &= \csc \theta - \cot \theta &= \pm\, \sqrt{1 - \cos \theta \over 1 + \cos \theta} &= \frac{\sin \theta}{1 + \cos \theta} &= \frac{1-\cos \theta}{\sin \theta} \end{align}and $\sin(\pi/2-\epsilon)=\cos(\epsilon)$ and $\cos(\pi/2-\epsilon)=\sin(\epsilon)$ in (1) yields
$$\boxed{ \begin{equation} \tan\left(\frac{\theta}{2}\right)= \frac{\cos(\epsilon)(1-\cos(\nu))}{\sin(\nu)(1+\sin(\epsilon))} \end{equation} }$$ ... a bit stuck now