\documentclass[a4paper,12pt]{article}
\newcommand\ds{\displaystyle}
\begin{document}

\parindent=0pt

QUESTION

If $\cosh x=2$ find $\sinh x$.

\bigskip


ANSWER

$\cosh^2x-\sinh^2x=1,\ \sinh^2x=\cosh^2x-1=2^2-1=3$

Therefore $\sinh x=\pm \sqrt{3}$





\end{document}
