Answer
$tanh(x + y)$ = $\frac{tanh(x)+tanh(y)}{1+tanh(x)tanh(y)}$
Work Step by Step
$tanh(x + y)$ = $\frac{sinh(x)}{cosh(x)}$
$tanh(x + y)$ = $\frac{e^{x+y}-e^{-(x+y)}}{e^{x+y}+e^{-(x+y)}}$
$tanh(x + y)$ = $\frac{2e^{x+y}-2e^{-(x+y)}}{2e^{x+y}+2e^{-(x+y)}}$
$tanh(x + y)$ = $\frac{e^{x+y}+e^{x-y}-e^{y-x}-e^{-(x+y)}+e^{x+y}+e^{y-x}-e^{x-y}-e^{-(x+y)}}{e^{x+y}+e^{x-y}+e^{y-x}+e^{-(x+y)}+e^{x+y}-e^{x-y}-e^{y-x}+e^{-(x+y)}}$
$tanh(x + y)$ = $\frac{(e^{x}-e^{-x})(e^{y}+e^{-y})+(e^{y}-e^{-y})(e^{x}+e^{-x})}{(e^{x}+e^{-x})(e^{y}+e^{-y})+(e^{x}-e^{-x})(e^{y}-e^{-y})}$
$tanh(x + y)$ = $\frac{\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}+\frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}}{1+\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}}$
$tanh(x + y)$ = $\frac{tanh(x)+tanh(y)}{1+tanh(x)tanh(y)}$
