prove-that-a-cosh-1-x-ln-x-x-2-1-b-tanh-1-x-1-2-ln-x-1-x-1-x-lt-1-

Question Number 178577 by Spillover last updated on 18/Oct/22

prove that

(a) \cosh^{- 1} x = \pm \ln (x + \sqrt{x^{2} - 1})

(b) \tanh^{- 1} x = \frac{1}{2} \ln (\frac{x + 1}{x - 1}), ∣ x ∣< 1

Answered by depressiveshrek last updated on 18/Oct/22

y = arccosh x

\cosh y = x

\frac{e^{y} + e^{- y}}{2} = x

e^{y} + \frac{1}{e^{y}} = 2 x

{(e^{y})}^{2} - 2 {x e}^{y} + 1 = 0

e^{y} = \frac{2 x + \sqrt{4 x^{2} - 4}}{2}^{*}

e^{y} = \frac{2 x + 2 \sqrt{x^{2} - 1}}{2}

e^{y} = x + \sqrt{x^{2} - 1}

y = \ln (x + \sqrt{x^{2} - 1})

arccosh x = \ln (x + \sqrt{x^{2} - 1})

y = arctanh x

\tanh y = x

\frac{e^{y} - e^{- y}}{e^{y} + e^{- y}} = x

e^{y} - \frac{1}{e^{y}} = {x e}^{y} + \frac{x}{e^{y}}

{(e^{y})}^{2} - x {(e^{y})}^{2} = x + 1

{(e^{y})}^{2} (1 - x) = x + 1

{(e^{y})}^{2} = \frac{x + 1}{1 - x}

e^{y} = \sqrt{\frac{x + 1}{1 - x}}^{*}

y = \ln \sqrt{\frac{x + 1}{1 - x}}

arctanh x = \frac{1}{2} \ln ∣ \frac{x + 1}{1 - x} ∣

Commented by Spillover last updated on 18/Oct/22

thank you