tanh ^{-1}left(frac{1}{2}right)+operatorname{ | vedclass

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(D) We know that $\operatorname{coth}^{-1}(x) = 	anh^{-1}\left(\frac{1}{x}ight)$ for $|x| > 1$. Therefore,$\operatorname{coth}^{-1}(2) = 	anh^{-1}

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$\log 3$

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(D) We know that $\operatorname{coth}^{-1}(x) = 	anh^{-1}\left(\frac{1}{x}ight)$ for $|x| > 1$. Therefore,$\operatorname{coth}^{-1}(2) = 	anh^{-1}\left(\frac{1}{2}ight)$. Substituting this into the expression,we get: $	anh ^{-1}\left(\frac{1}{2}ight)+\operatorname{coth}^{-1}(2) = 	anh ^{-1}\left(\frac{1}{2}ight)+	anh ^{-1}\left(\frac{1}{2}ight) = 2 	anh ^{-1}\left(\frac{1}{2}ight)$. Using the logarithmic form $	anh^{-1}(x) = \frac{1}{2} \log \left(\frac{1+x}{1-x}ight)$,we have: $2 	anh^{-1}\left(\frac{1}{2}ight) = 2 \cdot \frac{1}{2} \log \left(\frac{1+\frac{1}{2}}{1-\frac{1}{2}}ight) = \log \left(\frac{\frac{3}{2}}{\frac{1}{2}}ight) = \log 3$.

$\tanh ^{-1}\left(\frac{1}{2}\right)+\operatorname{coth}^{-1}(2)$ is equal to

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