Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 2 / b

For a real number $x$, the hyperbolic functions satisfy $\cosh x>0$ and $1-{\tanh }^{2}x={\mathrm{sech}}^{2}x$. Hence $\cosh x=1/\sqrt{1-u^2}$, and multiplying it by $\tanh x=u$ givesThe sign of the numerator preserves the sign of $\sinh x$.