operatorname{Tanh}^{-1}left(frac{1}{3}right)+ | vedclass

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Question

(D) We know that $\operatorname{Coth}^{-1}(x) = \operatorname{Tanh}^{-1}\left(\frac{1}{x}\right)$ for $|x| > 1$.
Given expression is $\operatornam

Vedclass · Accepted Answer

$\operatorname{Sinh}^{-1}\left(\frac{3}{4}\right)$

Vedclass · Answer

(D) We know that $\operatorname{Coth}^{-1}(x) = \operatorname{Tanh}^{-1}\left(\frac{1}{x}ight)$ for $|x| > 1$.
Given expression is $\operatorname{Tanh}^{-1}\left(\frac{1}{3}ight) + \operatorname{Coth}^{-1}(3)$.
Since $\operatorname{Coth}^{-1}(3) = \operatorname{Tanh}^{-1}\left(\frac{1}{3}ight)$,the expression becomes $\operatorname{Tanh}^{-1}\left(\frac{1}{3}ight) + \operatorname{Tanh}^{-1}\left(\frac{1}{3}ight) = 2 \operatorname{Tanh}^{-1}\left(\frac{1}{3}ight)$.
Using the logarithmic form $\operatorname{Tanh}^{-1}(x) = \frac{1}{2} \ln\left(\frac{1+x}{1-x}ight)$,we have:
$2 \operatorname{Tanh}^{-1}\left(\frac{1}{3}ight) = 2 	imes \frac{1}{2} \ln\left(\frac{1+1/3}{1-1/3}ight) = \ln\left(\frac{4/3}{2/3}ight) = \ln(2)$.
Since $\ln(2) = \operatorname{Sinh}^{-1}\left(\frac{3}{4}ight)$ (as $\operatorname{Sinh}^{-1}(x) = \ln(x + \sqrt{x^2+1})$ and $\ln(3/4 + \sqrt{9/16 + 1}) = \ln(3/4 + 5/4) = \ln(2)$),the correct option is $D$.

$\operatorname{Tanh}^{-1}\left(\frac{1}{3}\right)+\operatorname{Coth}^{-1}(3)=$

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