Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 2 / 13 / a

For $x>0$, the exponential definitions of the hyperbolic functions give $\cosh x-\sinh x=e^{-x}>0$ and $\sinh x>0$. Taking reciprocals of these positive quantities proves $1/\cosh x<1/\sinh x$. Also $\cosh x>1$ for $x>0$, so division by the positive $\sinh x$ gives $1/\sinh x<\cosh x/\sinh x$. Combining,