Answer
$$\tanh x = \frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}$$
Work Step by Step
$$\eqalign{
& {\text{using the definitions of the hyperbolic functions}}: \cr
& \cosh x = \frac{{{e^x} + {e^{ - x}}}}{2}{\text{ and }}\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2} \cr
& {\text{where the hyperbolic tangent is }}\tanh x = \frac{{\sinh x}}{{\cosh x}}.{\text{ then}} \cr
& \tanh x = \frac{{\sinh x}}{{\cosh x}} = \frac{{\frac{{{e^x} - {e^{ - x}}}}{2}}}{{\frac{{{e^x} + {e^{ - x}}}}{2}}} \cr
& \tanh x = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}} \cr
& {\text{multiplying the numerator and the denominator by }}{e^x} \cr
& \tanh x = \frac{{{e^x}\left( {{e^x} - {e^{ - x}}} \right)}}{{{e^x}\left( {{e^x} + {e^{ - x}}} \right)}} \cr
& {\text{simplifying}} \cr
& \tanh x = \frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}} \cr} $$
