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Answer 1

By the definition of the hyperbolic function tanh x, we have proven that $\frac{1-tanh \ x}{1 + tanh \ x}=e^{-2x}$

Hyperbolic functions & Proof of identities

By definition

$tanh \ x=\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}}$

Then,

$\frac{1-tanh \ x}{1 + tanh \ x}=\frac{1-\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} }{1+ \frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} }$

$=1-\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} \div (1+ \frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} })$

$=\frac{e^{x} +e^{-x}-(e^{x} -e^{-x})}{e^{x} +e^{-x}} \div (\frac{e^{x} +e^{-x}+e^{x} -e^{-x}}{e^{x} +e^{-x}} })$

$=\frac{e^{x} +e^{-x}-e^{x} +e^{-x}}{e^{x} +e^{-x}} \div \frac{e^{x} +e^{-x}+e^{x} -e^{-x}}{e^{x} +e^{-x}} }$

$=\frac{e^{-x}+e^{-x}}{e^{x} +e^{-x}} \div \frac{e^{x} +e^{x} }{e^{x} +e^{-x}} }$

$=\frac{2e^{-x}}{e^{x} +e^{-x}} \div \frac{2e^{x} }{e^{x} +e^{-x}} }$

$=\frac{2e^{-x}}{e^{x} +e^{-x}} \times \frac{e^{x} +e^{-x}}{2e^{x}}$

$=\frac{2e^{-x}}{1} \times \frac{1}{2e^{x}}$

$=\frac{2e^{-x}}{2e^{x}}$

$=\frac{e^{-x}}{e^{x}}$
$=e^{-x} \times \frac{1}{e^{x}}$

$= e^{-x} \times e^{-x}$

$= e^{-x+-x}$

$= e^{-x-x}$

$= e^{-2x}$

Hence, we have proven that $\frac{1-tanh \ x}{1 + tanh \ x}=e^{-2x}$

Learn more on Proof of Identities here: brainly.com/question/2561079

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