Sum of Squares of Hyperbolic Secant and Tangent

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Theorem

$\sech^2 x + \tanh^2 x = 1$

where $\sech$ and $\tanh$ are hyperbolic secant and hyperbolic tangent.


Proof

\(\ds \sech^2 x + \tanh^2 x\) \(=\) \(\ds \paren {\frac 2 {e^x + e^{-x} } }^2 + \tanh^2 x\) Definition 1 of Hyperbolic Secant
\(\ds \) \(=\) \(\ds \paren {\frac 2 {e^x + e^{-x} } }^2 + \paren {\frac {e^x - e^{-x} } {e^x + e^{-x} } }^2\) Definition 1 of Hyperbolic Tangent
\(\ds \) \(=\) \(\ds \frac {4 + e^{2 x} - 2 + e^{-2 x} } {e^{2 x} + 2 + e ^{-2 x} }\) Exponent Combination Laws
\(\ds \) \(=\) \(\ds \frac {e^{2 x} + 2 + e ^{-2 x} } {e^{2 x} + 2 + e ^{-2 x} }\)
\(\ds \) \(=\) \(\ds 1\)

$\blacksquare$


Also presented as

Sum of Squares of Hyperbolic Secant and Tangent can also be reported as:

$1 - \tanh^2 x = \sech^2 x$

or:

$1 - \sech^2 x = \tanh^2 x$


Also see


Sources