Answer
(a)$\qquad \displaystyle \frac{4}{3}$
(b)$\qquad \displaystyle \frac{13}{12}$
Work Step by Step
Definitions
$\displaystyle \sinh x=\frac{e^{x}-e^{-x}}{2} \qquad \mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}\ x=\displaystyle \frac{1}{\sinh x}=\frac{2}{e^{x}-e^{-x}},\quad x\neq 0$
$\displaystyle \cosh x=\frac{e^{x}+e^{-x}}{2} \qquad \mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}\ x=\displaystyle \frac{1}{\cosh x}=\frac{2}{e^{x}+e^{-x}}$
$\displaystyle \tanh x=\frac{\sinh x}{\cosh x} =\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \qquad \coth x=\frac{1}{\tanh x}=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}},\quad x\neq 0$
----
(a)
$\displaystyle \mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}\ (\ln 2)=\frac{2}{e^{\ln 2}-e^{-\ln 2}}\qquad $... $(e^{\ln x}=x)$ ... $(-\ln x=\ln x^{-1})$
$=\displaystyle \frac{2}{2-(\frac{1}{2})}=\frac{\frac{4}{2}}{\frac{3}{2}}=\frac{4}{3}$
(b)
$\displaystyle \coth(\ln 5)=\frac{\cosh(\ln 5)}{\sinh(\ln 5)}=\frac{e^{\ln 5}+e^{-\ln 5}}{e^{\mathrm{h}5}-e^{-\ln 5}}\qquad $
$=\displaystyle \frac{5+\frac{1}{5}}{5-(\frac{1}{5})}$
$=\displaystyle \frac{\frac{26}{5}}{\frac{24}{5}} =\frac{26}{24}=\frac{13}{12}$
