Prove-the-identity-cosh-1-x-sinh-1-1-x-please-help-

Question Number 5874 by sanusihammed last updated on 02/Jun/16

P r o v e t h e i d e n t i t y .

{c o s h}^{- 1} (x) = {s i n h}^{- 1} (\frac{1}{x})

p l e a s e h e l p .

Answered by Yozzii last updated on 02/Jun/16

{s i n h}^{- 1} u = l n (u + \sqrt{u^{2} + 1}) (u \in R)

L e t u = \frac{1}{x} (x \neq 0) .

\Rightarrow {s i n h}^{- 1} x^{- 1} = l n (\frac{1}{x} + \frac{1}{∣ x ∣} \sqrt{1 + x^{2}})

L e t x > 0 . \Rightarrow∣ x ∣= x .

\Rightarrow {s i n h}^{- 1} x^{- 1} = l n (\frac{1 + \sqrt{1 + x^{2}}}{x})

c o s h y = x

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

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

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

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

y = l n (x \pm \sqrt{x^{2} - 1})

{c o s h}^{- 1} x = l n (x \pm \sqrt{x^{2} - 1}) x ⩾ 1 .

L e t x = 1 \Rightarrow {c o s h}^{- 1} 1 = l n (1 \pm 0) = 0

B u t {s i n h}^{- 1} 1 = l n (\frac{1 + \sqrt{2}}{1}) \neq 0 = {c o s h}^{- 1} 1

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