How do you verify $tanh (x + y) = \frac{tanh x + tanh y}{1 + tanh x + tanh y}$ $tanh(x + y) = (tanh x + tanh y)/(1 + tanh x + tanh y)$ ?

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How do you verify $tanh (x + y) = \frac{tanh x + tanh y}{1 + tanh x + tanh y}$ $tanh(x + y) = (tanh x + tanh y)/(1 + tanh x + tanh y)$ ?

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Answer 1 · 2018-07-31T10:43:50

Please see the explanation bellow

Explanation:

You need

$sinh (x + y) = sinh x cosh y + cosh x sinh y$ $sinh(x+y)=sinhxcoshy+coshxsinhy$

$cosh (x + y) = cosh x cosh y + sinh x sinh y$ $cosh(x+y)=coshxcoshy+sinhxsinhy$

You can either start with

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

Or with

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

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

Dividing all the terms by $cosh (x) cosh (y)$ $cosh(x)cosh(y)$

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

$= \frac{tanh x + tanh y}{1 + tanh x tanh y}$ $=(tanhx+tanhy)/(1+tanhxtanhy)$

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How do you verify $tanh (x + y) = \frac{tanh x + tanh y}{1 + tanh x + tanh y}$ $tanh(x + y) = (tanh x + tanh y)/(1 + tanh x + tanh y)$ ?

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Explanation:

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How do you verify $tanh (x + y) = \frac{tanh x + tanh y}{1 + tanh x + tanh y}$ $tanh(x + y) = (tanh x + tanh y)/(1 + tanh x + tanh y)$ ?