Prove that, can anyone help me, please?

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Prove that, can anyone help me, please?

$cosh [ln (x +$ $cosh[ln(x+$ $\sqrt{(x^{2} - 1)}]$ $sqrt((x^2-1)]]$ = $x$ $x$

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Answer 1 · 2017-12-11T02:14:12

By definition:

$cosh x = \frac{e^{x} + e^{- x}}{2}$ $coshx =(e^x+e^-x)/2$

so:

$cosh ln (x + \sqrt{x^{2} - 1}) = \frac{1}{2} (e^{ln (x + \sqrt{x^{2} - 1})} + e^{- ln (x + \sqrt{x^{2} - 1})})$ $coshln(x+sqrt(x^2-1)) = 1/2(e^(ln(x+sqrt(x^2-1)))+e^(-ln(x+sqrt(x^2-1))))$

$cosh ln (x + \sqrt{x^{2} - 1}) = \frac{1}{2} (x + \sqrt{x^{2} - 1} + \frac{1}{x + \sqrt{x^{2} - 1}})$ $coshln(x+sqrt(x^2-1)) = 1/2(x+sqrt(x^2-1)+1/(x+sqrt(x^2-1)))$

$cosh ln (x + \sqrt{x^{2} - 1}) = \frac{1}{2} \frac{{(x + \sqrt{x^{2} - 1})}^{2} + 1}{x + \sqrt{x^{2} - 1}}$ $coshln(x+sqrt(x^2-1)) = 1/2((x+sqrt(x^2-1))^2+1)/(x+sqrt(x^2-1))$

$cosh ln (x + \sqrt{x^{2} - 1}) = \frac{1}{2} (\frac{x^{2} + 2 x \sqrt{x^{2} - 1} + x^{2} - 1 + 1}{x + \sqrt{x^{2} - 1}})$ $coshln(x+sqrt(x^2-1)) = 1/2((x^2+2xsqrt(x^2-1)+x^2-1+1)/(x+sqrt(x^2-1)))$

$cosh ln (x + \sqrt{x^{2} - 1}) = \frac{1}{2} (\frac{2 x^{2} + 2 x \sqrt{x^{2} - 1}}{x + \sqrt{x^{2} - 1}})$ $coshln(x+sqrt(x^2-1)) = 1/2((2x^2+2xsqrt(x^2-1))/(x+sqrt(x^2-1)))$

$cosh ln (x + \sqrt{x^{2} - 1}) = x \frac{x + \sqrt{x^{2} - 1}}{x + \sqrt{x^{2} - 1}}$ $coshln(x+sqrt(x^2-1)) =x(x+sqrt(x^2-1))/(x+sqrt(x^2-1))$

$cosh ln (x + \sqrt{x^{2} - 1}) = x$ $coshln(x+sqrt(x^2-1)) =x$

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Prove that, can anyone help me, please?

$cosh [ln (x +$ $cosh[ln(x+$ $\sqrt{(x^{2} - 1)}]$ $sqrt((x^2-1)]]$ = $x$ $x$

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Prove that, can anyone help me, please?

cosh[ln(x+cosh[ln(x+√(x2−1)]sqrt((x^2-1)]] = xx

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$cosh [ln (x +$ $cosh[ln(x+$ $\sqrt{(x^{2} - 1)}]$ $sqrt((x^2-1)]]$ = $x$ $x$