Rearrangement of one equation involving angles I am reading an astrodynamics textbook and I found the following system of equations:
$$e\cdot \cos(v_1)=p/r_1 - 1;$$
$$e\cdot \cos(v_2)=p/r_2 - 1;$$
$e$ refers to the eccentricity of the conic section, $p$ the semi-latus rectum, $v_1$ and $v_2$ are angles for times $t_1$ and $t_2$ and $r_1$ and $r_2$ are the absolute value of radius vectors at times $t_1$ and $t_2$. Since there are more unknowns ($e$, $v_1$ and $v_2$. It is assumed that $p$ is known) than equations the angle $v_2$  is expressed as $v_1$ plus a $\Delta V$. From there the second equation changes to:
$$e\cdot \sin(v_1)=(-p/r_2+1+(p/r_2-1)\cos(\Delta V))\csc(\Delta V)$$
I was wondering if someone could explain the intermediary steps plus the fact that the differential in the angle is taken by cosecant of this $\Delta V$.
Thank you very much for your time.
J
 A: To make the notation simpler I will call
\begin{eqnarray}
c_1 &=& \cos v_1 = \frac{1}{e}\left(\frac{p}{r_1} - 1\right) \\
c_2 &=& \cos v_2 = \frac{1}{e}\left(\frac{p}{r_2} - 1\right) \\
\delta &=& \sin(v_2 - v_1) \\
\Delta &=& \cos(v_2 - v_1)
\end{eqnarray}
With this in mind,
\begin{eqnarray}
\sin(v_2 - v_1) &=& \sin v_2\cos v_1 - \sin v_1\cos v_2 \\
\delta &=&  \color{red}{c_1\sin v_2} - c_2 \sin v_1 \tag{1}
\end{eqnarray}
and
\begin{eqnarray}
\cos (v_2 - v_1) &=& \cos v_2 \cos v_1 + \sin v_2 \sin v_1 \\
\Delta &=& c_1 c_2 + \sin v_2 \sin v_1 \tag{2}
\end{eqnarray}
Now it is a matter of playing with these two equations
\begin{eqnarray}
c_1 \Delta &=& c_1^2 c_2 + \sin v_1 (\color{red}{c_1 \sin v_2}) \\
c_1\Delta&\stackrel{(1)}{=}& c_1^2 c_2 + \sin v_1(\color{red}{\delta + c_2 \sin v_1}) \\
c_1\Delta &=& c_1^2c_2  + \delta\sin v_1 + c_2\color{blue}{\sin^2v_1} \\
c_1\Delta &=& c_1^2c_2  + \delta\sin v_1 + c_2(\color{blue}{1 - \cos^2v_1}) \\
c_1\Delta &=& c_1^2c_2  + \delta\sin v_1 + c_2(\color{blue}{1 - c_1^2}) \\
c_1\Delta &=& c_2 + \delta \sin v_1 \tag{3}
\end{eqnarray}
And from here just solve for $\sin v_1$
\begin{eqnarray}
\sin v_1 &=& \frac{1}{\delta}\left(c_1\Delta - c_2 \right) \\
\sin v_1 &=& \frac{1}{\sin(v_2 - v_1)} \left(\cos v_1 \cos(v_2 - v_1) - \cos v_2 \right) \\
\sin v_1 &=& \csc(v_2 - v_1) \left[\frac{1}{e}\left(\frac{p}{r_1} - 1\right)\cos (v_2 - v_1) - \frac{1}{e}\left(\frac{p}{r_2} - 1\right) \right] \tag{4}
\end{eqnarray}
