I Trig Manipulations I'm Not Getting

[thatboi]

Hi all,
I am starting with the following equation: $2\cot\left(\frac{\theta}{2}\right) = \cot\left(\frac{k_{1}}{2}\right) - \cot\left(\frac{k_{2}}{2}\right)$
with the following definitions: $k_{1} = \frac{K}{2} + ik, k_{2} = \frac{K}{2}-ik, \theta = \pi(I_{2}-I_{1}) + iNk$, where $k,K,N\in\mathbb{R}$ and $I_{2},I_{1}\in\mathbb{Z}$. I wish to plug these definitions into the above equation and get the new equation: $\cos\left(\frac{K}{2}\right)\sinh(Nk)=\sinh[(N-1)k]+\cos[\pi(I_{2}-I_{1})]\sinh(k)$. I have done the following (first use the identity $\cot\frac{\theta}{2} = \frac{1+\cos\theta}{\sin\theta}$):
\begin{equation}
\begin{split}
&2\frac{1+\cos\theta}{\sin\theta} = \frac{1+\cos k_{1}}{\sin k_{1}}-\frac{1+\cos k_{2}}{\sin k_{2}} \\
&\rightarrow 2\frac{1+\cos(\pi(I_{2}-I_{1}))\cos(iNk)-\sin(\pi(I_{2}-I_{1}))\sin(iNk)}{\sin(\pi(I_{2}-I_{1}))\cos(iNk)+\sin(iNk)\cos(\pi(I_{2}-I_{1}))} = \frac{1+\cos\frac{K}{2}\cos(ik)-\sin\frac{K}{2}\sin(ik)}{\sin\frac{K}{2}\cos(ik)+\cos\frac{K}{2}\sin(ik)}-\frac{1+\cos\frac{K}{2}\cos(ik)+\sin\frac{K}{2}\sin(ik)}{\sin\frac{K}{2}\cos(ik)-\cos\frac{K}{2}\sin(ik)}\\
&\rightarrow -2i\frac{1+\cos(\pi(I_{2}-I_{1}))\cosh(Nk)}{\sinh(Nk)\cos(\pi(I_{2}-I_{1}))} = \frac{2i\sinh(k)}{\cos\left(\frac{K}{2}\right)-\cosh(k)}
\end{split}
\end{equation}
where I dropped terms with $\sin(\pi(I_{2}-I_{1}))$ since they would evaluate to 0. While this is a lot more simplified I still don't know how to get to the final form of equation I want or if I made an error somewhere. Any advice appreciated!

 

[Charles Link]

The formula that you start with in your first line looks very unrealistic, when following the definitions you give for $\theta$ and the various k's. It looks to me like the original problem statement may contain an error or two.