If 2 operatorname{Tanh}^{-1} x = operatorname | vedclass

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(C) Given $2 \operatorname{Tanh}^{-1} x = \operatorname{Sinh}^{-1}\left(\frac{4}{3}ight)$.Let $\operatorname{Tanh}^{-1} x = 	heta$, then $x = 	anh

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$\log (2+\sqrt{3})$

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(C) Given $2 \operatorname{Tanh}^{-1} x = \operatorname{Sinh}^{-1}\left(\frac{4}{3}ight)$.Let $\operatorname{Tanh}^{-1} x = 	heta$, then $x = 	anh 	heta$.So, $2	heta = \operatorname{Sinh}^{-1}\left(\frac{4}{3}ight)$, which implies $\sinh(2	heta) = \frac{4}{3}$.Using the identity $\sinh(2	heta) = \frac{2	anh 	heta}{1-	anh^2 	heta}$, we have $\frac{2x}{1-x^2} = \frac{4}{3}$.$6x = 4 - 4x^2 \implies 4x^2 + 6x - 4 = 0 \implies 2x^2 + 3x - 2 = 0$.Solving for $x$: $(2x-1)(x+2) = 0$. Since $|x| Now, we need to find $\operatorname{Cosh}^{-1}\left(\frac{1}{x}ight) = \operatorname{Cosh}^{-1}(2)$.Using the formula $\operatorname{Cosh}^{-1} y = \log(y + \sqrt{y^2-1})$, we get $\operatorname{Cosh}^{-1}(2) = \log(2 + \sqrt{2^2-1}) = \log(2 + \sqrt{3})$.

If $2 \operatorname{Tanh}^{-1} x = \operatorname{Sinh}^{-1}\left(\frac{4}{3}\right)$, then $\operatorname{Cosh}^{-1}\left(\frac{1}{x}\right) = $

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