Double Angle Formulas/Hyperbolic Tangent/Proof 3

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Theorem

$\tanh 2 x = \dfrac {2 \tanh x} {1 + \tanh^2 x}$


Proof

Starting from the right, we have

\(\ds \dfrac {2 \tanh x} {1 + \tanh^2 x}\) \(=\) \(\ds \dfrac {2 \paren {\dfrac {e^x - e^{-x} } {e^x + e^{-x} } } } {1 + \paren {\dfrac{e^x - e^{-x} } {e^x + e^{-x} } }^2}\) Definition 1 of Hyperbolic Tangent
\(\ds \) \(=\) \(\ds \dfrac {2 \paren {e^x + e^{-x} } \paren {e^x - e^{-x} } } {\paren {e^x + e^{-x} }^2 + \paren {e^x - e^{-x} }^2}\) multiplying both numerator and denominator by $\paren {e^x + e^{-x} }^2$
\(\ds \) \(=\) \(\ds \dfrac {2 \paren {e^{2 x} - e^{-2 x} } } {e^{2 x} + 2 + e^{-2 x} + e^{2 x} - 2 + e^{-2 x} }\) Difference of Two Squares, Square of Sum, Square of Difference
\(\ds \) \(=\) \(\ds \dfrac {2 \paren {e^{2 x} - e^{-2 x} } } {2 e^{2 x} + 2 e^{-2 x} }\) simplifying
\(\ds \) \(=\) \(\ds \dfrac {e^{2 x} - e^{-2 x} } {e^{2 x} + e^{-2 x} }\) simplifying
\(\ds \) \(=\) \(\ds \tanh {2 x}\) Definition 1 of Hyperbolic Tangent

$\blacksquare$