How do you verify tanh(x + y) = (tanh x + tanh y)/(1 + tanh x + tanh y)?

Jul 31, 2018

Explanation:

You need

$\sinh \left(x + y\right) = \sinh x \cosh y + \cosh x \sinh y$

$\cosh \left(x + y\right) = \cosh x \cosh y + \sinh x \sinh y$

$\tanh \left(x + y\right) = \frac{{e}^{x + y} - {e}^{- x - y}}{{e}^{x + y} + {e}^{- x - y}}$

Or with

$\tanh \left(x + y\right) = \sinh \frac{x + y}{\cosh} \left(x + y\right)$

$= \frac{\sinh \left(x\right) \cosh \left(y\right) + \sinh \left(y\right) \cosh \left(x\right)}{\cosh \left(x\right) \cosh \left(y\right) + \sinh \left(x\right) \sinh \left(y\right)}$

Dividing all the terms by $\cosh \left(x\right) \cosh \left(y\right)$

$= \frac{\frac{\sinh \left(x\right) \cosh \left(y\right)}{\cosh \left(x\right) \cosh \left(y\right)} + \frac{\sinh \left(y\right) \cosh \left(x\right)}{\cosh \left(x\right) \cosh \left(y\right)}}{\frac{\cosh \left(x\right) \cosh \left(y\right)}{\cosh \left(x\right) \cosh \left(y\right)} + \frac{\sinh \left(x\right) \sinh \left(y\right)}{\cosh \left(x\right) \cosh \left(y\right)}}$

$= \frac{\tanh x + \tanh y}{1 + \tanh x \tanh y}$