# How can you prove that 1 - tanh^2x = sech^2x?

Feb 7, 2017

Let's rewrite in terms of ${e}^{x}$, using the identities $\sech x = \frac{2}{{e}^{x} + {e}^{-} x}$ and $\tanh x = \frac{{e}^{x} - {e}^{-} x}{{e}^{x} + {e}^{-} x}$.

$1 - {\left(\frac{{e}^{x} - {e}^{-} x}{{e}^{x} + {e}^{-} x}\right)}^{2} = {\left(\frac{2}{{e}^{x} + {e}^{-} x}\right)}^{2}$

$1 - \frac{\left({e}^{x} - {e}^{-} x\right) \left({e}^{x} - {e}^{-} x\right)}{\left({e}^{x} + {e}^{-} x\right) \left({e}^{x} + {e}^{-} x\right)} = \left(\frac{2}{{e}^{x} + {e}^{-} x}\right) \left(\frac{2}{{e}^{x} + {e}^{-} x}\right)$

$1 - \frac{{e}^{x} {e}^{x} - {e}^{-} x {e}^{x} - {e}^{-} x {e}^{x} + {e}^{-} x {e}^{-} x}{{e}^{x} {e}^{x} + {e}^{-} x {e}^{x} + {e}^{-} x {e}^{x} + {e}^{-} x {e}^{-} x} = \frac{4}{{e}^{x} {e}^{x} + {e}^{-} x {e}^{x} + {e}^{-} x {e}^{x} + {e}^{-} x {e}^{-} x}$

$1 - \frac{{e}^{2 x} - {e}^{0} - {e}^{0} + {e}^{- 2 x}}{{e}^{2 x} + {e}^{0} + {e}^{0} + {e}^{- 2 x}} = \frac{4}{{e}^{2 x} + {e}^{0} + {e}^{0} + {e}^{- 2 x}}$

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

$\frac{{e}^{2 x} + 2 + {e}^{- 2 x} - {e}^{2 x} + 2 - {e}^{- 2 x}}{{e}^{2 x} + 2 + {e}^{- 2 x}} = \frac{4}{{e}^{2 x} + 2 + {e}^{- 2 x}}$

$\frac{4}{{e}^{2 x} + 2 + {e}^{- 2 x}} = \frac{4}{{e}^{2 x} + 2 + {e}^{- 2 x}}$

This has been proved!

Hopefully this helps!