tanh^{-1}left(frac{1}{3}right) + coth^{-1}(2) | vedclass

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(A) We use the logarithmic definitions of inverse hyperbolic functions: $	anh^{-1}(x) = \frac{1}{2} \log \left(\frac{1+x}{1-x}ight)$ for $|x| < 1$

Vedclass · Accepted Answer

$\log \sqrt{6}$

Vedclass · Answer

(A) We use the logarithmic definitions of inverse hyperbolic functions: $	anh^{-1}(x) = \frac{1}{2} \log \left(\frac{1+x}{1-x}ight)$ for $|x|  1$. Substituting $x = \frac{1}{3}$ in $	anh^{-1}(x)$: $	anh^{-1}\left(\frac{1}{3}ight) = \frac{1}{2} \log \left(\frac{1 + 1/3}{1 - 1/3}ight) = \frac{1}{2} \log \left(\frac{4/3}{2/3}ight) = \frac{1}{2} \log(2)$. Substituting $x = 2$ in $\coth^{-1}(x)$: $\coth^{-1}(2) = \frac{1}{2} \log \left(\frac{2+1}{2-1}ight) = \frac{1}{2} \log(3)$. Adding these results: $\frac{1}{2} \log(2) + \frac{1}{2} \log(3) = \frac{1}{2} \log(2 	imes 3) = \frac{1}{2} \log(6) = \log(6^{1/2}) = \log \sqrt{6}$.

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

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