Proof That csch^2(x) = coth^2(x) -1

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

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

If we square this:

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

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

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

If we square this:

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

If we do \(\coth^2(x) - \operatorname{csch}^2(x)\):

\[ \frac{e^{2x} +2+ e^{-2x}}{e^{2x} -2+ e^{-2x}} - \frac{4}{e^{2x} - 2 + e^{-2x}} = \frac{e^{2x} -2+ e^{-2x}}{e^{2x} -2+ e^{-2x}} = 1 \]

This means \(\operatorname{csch}^2(x) = \coth^2(x) - 1 \)

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