\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
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\begin{document}

{\bf Question}

State and prove the hyperbolic Pythagorean theorem, relating the
hyperbolic lengths of the three sides of a hyperbolic right
triangle.

\medskip
\noindent Use this hyperbolic Pythagorean theorem to rework
Exercise 3, Sheet 8.
\medskip

{\bf Answer} %Note ref to ex 3 sheet 8%

\begin{center}
$\begin{array}{c}
\epsfig{file=407-9-4.eps, width=40mm}
\end{array}
\ \ \
\begin{array}{rl}
\cosh(c) = & \cosh(a)\cosh(b)\\
 & -\sinh(a)\sinh(b)\cos(\gamma)\\
= & \cosh(a)\cosh(b)\\
 & \un{\textrm{from lcI}}
\end{array} $
\end{center}

\un{Exercise 3, sheet 8}

\begin{center}
\epsfig{file=407-9-5.eps, width=60mm}
\end{center}

$a=\ln(5)$
\begin{eqnarray*} \cosh(a) & = & \ds\frac{1}{2}\left(e^{\ln(5)}+e^{-\ln(5)}\right)\\ &
= & \ds\frac{1}{2}\left(5+\ds\frac{1}{5}\right)=\ds\frac{13}{5}
\end{eqnarray*}

\begin{eqnarray*} b=\ds\int_{\theta}^{\frac{\pi}{2}}
\ds\frac{1}{\sin(t)} \,dt & = & -\ln|\csc(\theta)-\cot(\theta)|\\
& = & \ln \left|\ds\frac{\sin(\theta)}{1-\cos(\theta)} \right|
\end{eqnarray*}

\begin{eqnarray*} \cosh(b)=\ds\frac{1}{2}(e^b+e^{-b})\\ & = &
\ds\frac{1}{2}\left(\ds\frac{\sin(\theta)}{1-\cos(\theta)}+
\ds\frac{1-\cos(\theta)}{\sin(\theta)}\right)\\ & = &
\ds\frac{1}{2}\left(\ds\frac{\sin^2(\theta)+(1-\cos(\theta)^2}
{\sin(\theta)(1-\cos(\theta))}\right)\\ & = &
\ds\frac{\sin^2(\theta)+1-2\cos(\theta)+\cos^2(\theta)}{2\sin(\theta)(1-\cos(\theta))}\\
& = & \ds\frac{1}{\sin(\theta)}\\ & = & \un{\csc(\theta)}
\end{eqnarray*}

so $\cosh(c)=\cosh(a)\cos(b)=\ds\frac{13}{5}\csc(\theta)$

$e^c+e^{-c}=\ds\frac{26}{5}\csc(\theta)$

$e^{2c}-\ds\frac{26}{5}\csc(\theta)e^c+1=0$

$\begin{array} {rcl} e^c & = &
\ds\frac{1}{2}\left[\ds\frac{26}{5}\csc(\theta)+\sqrt{\left(\ds\frac{26}{5}\right)
^2\csc^2(\theta)-4}\right]\\ {} & = &
\ds\frac{13}{5}\csc(\theta)+\sqrt{\left(\ds\frac{13}{5}\right)^2\csc^2(\theta)-1}\\
c & = &
\un{\ln\left(\ds\frac{13}{5}\csc(\theta)+\sqrt{\left(\ds\frac{13}{5}\right)^2
\csc^2(\theta)-1}\right)} \end{array}$

\end{document}