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Prove that cosh2x – sinh2x = 1

Prove that

    \[ \cosh^2 x - \sinh^2 x = 1. \]


Proof. We use the definitions of \cosh x and \sinh x in terms of the exponential function,

    \begin{align*}  \cosh^2 x - \sinh^2 x &= \left( \frac{e^x + e^{-x}}{2} \right)^2 - \left( \frac{e^x - e^{-x}}{2} \right)^2 \\  &= \frac{e^{2x} + 2 + e^{-2x} - e^{2x} + 2 - e^{-2x}}{4} \\  &= \frac{4}{4} \\  &= 1. \qquad \blacksquare \end{align*}

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