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{2}+\operatorname{coth}^{-1} 3=$
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2876 Upvotes
Verified Answer
The correct answer is:
$\log \sqrt{6}$
$$
\begin{aligned}
& \text {} \tanh ^{-1}\left(\frac{1}{2}\right)+\operatorname{coth} h^{-1} \text { (3) } \\
& =\frac{1}{2} \ln \left(\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\right)+\frac{1}{2} \ln \left(\frac{3+1}{3-1}\right)=\frac{1}{2} \log \left(\frac{\frac{3}{2}}{\frac{1}{2}}\right)+\frac{1}{2} \log \left(\frac{4}{2}\right) \\
& =\frac{1}{2} \log 3+\frac{1}{2} \log 2 \\
& =\log \sqrt{3}+\log \sqrt{2} \\
& =\log (\sqrt{3} \cdot \sqrt{2})=\log \sqrt{6} \text {. } \\
\end{aligned}
$$
\begin{aligned}
& \text {} \tanh ^{-1}\left(\frac{1}{2}\right)+\operatorname{coth} h^{-1} \text { (3) } \\
& =\frac{1}{2} \ln \left(\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\right)+\frac{1}{2} \ln \left(\frac{3+1}{3-1}\right)=\frac{1}{2} \log \left(\frac{\frac{3}{2}}{\frac{1}{2}}\right)+\frac{1}{2} \log \left(\frac{4}{2}\right) \\
& =\frac{1}{2} \log 3+\frac{1}{2} \log 2 \\
& =\log \sqrt{3}+\log \sqrt{2} \\
& =\log (\sqrt{3} \cdot \sqrt{2})=\log \sqrt{6} \text {. } \\
\end{aligned}
$$
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