## Calculus (3rd Edition)

$$\cosh x=\frac{\sqrt{41}}{5}\quad \tanh x= \frac{4}{\sqrt{41}} .$$
Since, $\sinh x=0.8=\frac{4}{5}$, then using the property $$\cosh^2x-\sinh^2x=1\Longrightarrow \cosh x=\frac{\sqrt{41}}{5}.$$ Now, $$\tanh x=\frac{\sinh x}{\cosh x}=\frac{4}{\sqrt{41}}.$$