Proof That sech^2(x) = 1 - tanh^2(x)

\(\operatorname{sech}(x)\) is defined as:

\[\operatorname{sech}(x) = \frac{2}{e^x + e^{-x}} \]

If we square this:

\[\operatorname{sech}^2(x) = \frac{4}{e^{2x} + 2 + e^{-2x}} \]

\(\tanh(x)\) is defiend as:

\[\tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} \]

If we square this:

\[ \tanh^2(x) = \frac{e^{2x} -2+ e^{-2x}}{e^{2x} +2+ e^{-2x}} \]

If we add \(\tanh^2(x)\) and \(\operatorname{sech}^2(x)\):

\[ \tanh^2(x) + \operatorname{sech}^2(x) = \frac{e^{2x} -2+ e^{-2x}}{e^{2x} +2+ e^{-2x}} + \frac{4}{e^{2x} + 2 + e^{-2x}} = \frac{e^{2x} +2+ e^{-2x}}{e^{2x} +2+ e^{-2x}} = 1 \]

This means \(\operatorname{sech}^2(x) = 1 - \tanh^2(x)\)

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