Answer
a) 0
b) $\dfrac{e^2-1}{e^2+1}$
Work Step by Step
Always remember
$\tanh = \dfrac{\sinh x}{\cosh x}
=\dfrac{\dfrac{e^x-e^{-x}}{2}}{\dfrac{e^x+e^{-x}}{2}}
=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}$
a) solve $\tanh 0$
$=\dfrac{e^0-e^{-0}}{e^0+e^{-0}}$
or ,$=\frac{1-1}{1+1}$
or, $=0$
b) solve $\tanh1$
$=\dfrac{e^1-e^{-1}}{e^1+e^{-1}}$
or, $=\dfrac{e-\frac{1}{e}}{e+\frac{1}{e}}$
or, $=\frac{e^2-1}{e^2+1}$
