## Thomas' Calculus 13th Edition

$\sinh x=\dfrac{4}{3}\\ \cosh x= \dfrac{5}{3} ;\\ \tanh x= \dfrac{4}{5}; \\ sech x=\dfrac{3}{5} ; \\ csch x=\dfrac{3}{4}$ and $coth x=\dfrac{5}{4}$
Given: $\sinh x=\dfrac{4}{3}$ The remaining hyperbolic functions can be found as follows: Use the fact $\cosh^2 x-\sinh^2x=1$ or, $\cosh^2 x-(\dfrac{4}{3})^2=1 \\ or, \cosh^2 x=\dfrac{25}{9}$ Thus, $\cosh x= \dfrac{5}{3} \\ \tanh x= \dfrac{\sin hx}{\cosh x}=\dfrac{\dfrac{4}{3}}{\dfrac{5}{3}}=\dfrac{4}{5}$ Now, $sech x= \dfrac{1}{\cosh x}=\dfrac{1}{\dfrac{5}{3}}=\dfrac{3}{5} \\ csch x= \dfrac{1}{\sinh x}=\dfrac{1}{\dfrac{4}{3}}=\dfrac{3}{4}$ and $coth x= \dfrac{1}{\tanh x}=\dfrac{1}{\dfrac{4}{5}}=\dfrac{5}{4}$