Find the exact value of:

Solution

(a) $\cosh 0=\dfrac{e^{0}+e^{0}}{2}=\dfrac{2}{2}=1$

(b) ${\mathop{\rm{sech}}}~0=\dfrac{1}{\cosh 0}=\dfrac{1}{1}=1$

(c) $\tanh ( \ln 2) =\dfrac{e^{\ln 2}-e^{-\ln 2}}{e^{\ln 2}+e^{-\ln 2}}=\dfrac{2-e^{\ln ( 1/2) }}{2+e^{\ln ( 1/2) }}=\dfrac{2-\dfrac{1}{2}}{2+\dfrac{1}{2}}=\dfrac{3}{5}$