Answer
See below
Work Step by Step
By using property P9 we obtain:
$$\begin{array}{c}{\det (A)=\left[\begin{array}{ccc}
\cosh x & \sin x\\ \sinh x & \cosh x \end{array}\right] }\end{array} =\cosh^ 2x-\sinh^2 x=1\\
\begin{array}{c}{\det (A)=\left[\begin{array}{ccc}
\cosh y & \sinh y\\ \sinh y & \cosh y \end{array}\right] }\end{array} =\cosh^ 2y-\sinh^2 y=1$$
then $\begin{array}{c}{\det (A)=\left[\begin{array}{ccc}
\cosh x\cosh y+ \sinh x\sinh y &\cosh x \sinh y+\sinh x\cosh y\\ \sinh x\cosh y+\cosh x\sinh y & \sinh x\sinh y+\cosh x \cosh y\end{array}\right] }\end{array} =[(\cosh x\cosh y+ \sinh x\sinh y).( \sinh x\sinh y+\cosh x\cosh y)]-[(\cosh x \sinh y+\sinh x\cosh y)(\sinh x\cosh y+\cosh x\sinh y)]\\=(\cosh (x+y).\cosh (x+y))-(\sinh (x+y).\sinh(x+y))\\
=\cosh^2(x+y)-\sinh^2 (x+y)\\=1$
Consequently, $\det(AB)=\det(A).\det(B).$
