Question

Prove that -Cos h2x=1+tan h^2x/1-tanh^2x

Answer 1

Step-by-step explanation:

We have

$\dfrac{1 + \tanh^{2} (x)}{1 - \tanh^{2} (x) }$

$= \dfrac{1 + \left \{ \dfrac{ {e}^{x} - {e}^{ - x} }{{e}^{x} + {e}^{ - x} } \right \}^{2} }{1 - \left \{ \dfrac{ {e}^{x} - {e}^{ - x} }{{e}^{x} + {e}^{ - x} } \right \}^{2} }$

$= \dfrac{1 + \dfrac{ ({e}^{x} - {e}^{ - x} )^{2} }{({e}^{x} + {e}^{ - x} )^{2} } }{1 - \dfrac{ ({e}^{x} - {e}^{ - x} )^{2} }{({e}^{x} + {e}^{ - x} )^{2} } }$

$= \dfrac{ \dfrac{({e}^{x} + {e}^{ - x} )^{2} + ({e}^{x} - {e}^{ - x} )^{2} }{({e}^{x} + {e}^{ - x} )^{2} } }{ \dfrac{({e}^{x} + {e}^{ - x} )^{2} - ({e}^{x} - {e}^{ - x} )^{2} }{({e}^{x} + {e}^{ - x} )^{2} } }$

$= \dfrac{ ({e}^{x} + {e}^{ - x} )^{2} + ({e}^{x} - {e}^{ - x} )^{2} }{ ({e}^{x} + {e}^{ - x} )^{2} - ({e}^{x} - {e}^{ - x} )^{2} }$

$= \dfrac{ 2 \left\{({e}^{x}) ^{2} + ({e}^{ - x} )^{2} \right\} }{ 4 \cdot{e}^{x} \cdot {e}^{ - x} }$

$= \dfrac{ {e}^{2x} + {e}^{ - 2x} }{ 2 \cdot{e}^{x} \cdot {e}^{ - x} }$

$= \dfrac{ {e}^{2x} + {e}^{ - 2x} }{ 2 }$

$= \cosh(2x)$