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

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(A) We use the logarithmic definitions of inverse hyperbolic functions: $	anh^{-1}(x) = \frac{1}{2} \ln(\frac{1+x}{1-x})$ and $\operatorname{coth}^{-

Vedclass · Accepted Answer

$\log \sqrt{6}$

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(A) We use the logarithmic definitions of inverse hyperbolic functions: $	anh^{-1}(x) = \frac{1}{2} \ln(\frac{1+x}{1-x})$ and $\operatorname{coth}^{-1}(x) = \frac{1}{2} \ln(\frac{x+1}{x-1})$.For $	anh^{-1}(\frac{1}{2})$:$	anh^{-1}(\frac{1}{2}) = \frac{1}{2} \ln(\frac{1+1/2}{1-1/2}) = \frac{1}{2} \ln(\frac{3/2}{1/2}) = \frac{1}{2} \ln(3) = \log \sqrt{3}$.For $\operatorname{coth}^{-1}(3)$:$\operatorname{coth}^{-1}(3) = \frac{1}{2} \ln(\frac{3+1}{3-1}) = \frac{1}{2} \ln(\frac{4}{2}) = \frac{1}{2} \ln(2) = \log \sqrt{2}$.Adding these results:$\log \sqrt{3} + \log \sqrt{2} = \log(\sqrt{3} \cdot \sqrt{2}) = \log \sqrt{6}$.

$\tanh^{-1}(\frac{1}{2}) + \operatorname{coth}^{-1}(3) = $

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