operatorname{coth}^{-1} 3 + tanh^{-1} frac{1} | vedclass

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Question

(B) We use the logarithmic definitions of inverse hyperbolic functions: $\operatorname{coth}^{-1} x = \frac{1}{2} \log_e \left(\frac{x+1}{x-1}\right)$

Vedclass · Accepted Answer

$\log_e 2\sqrt{3}$

Vedclass · Answer

(B) We use the logarithmic definitions of inverse hyperbolic functions: $\operatorname{coth}^{-1} x = \frac{1}{2} \log_e \left(\frac{x+1}{x-1}ight)$ for $|x| > 1$. $	anh^{-1} x = \frac{1}{2} \log_e \left(\frac{1+x}{1-x}ight)$ for $|x| $\operatorname{cosech}^{-1} x = \log_e \left(\frac{1 + \sqrt{1+x^2}}{x}ight)$ for $x > 0$ and $\operatorname{cosech}^{-1} x = \log_e \left(\frac{1 - \sqrt{1+x^2}}{x}ight)$ for $x Step $1$: Calculate each term. $\operatorname{coth}^{-1}(3) = \frac{1}{2} \log_e \left(\frac{3+1}{3-1}ight) = \frac{1}{2} \log_e(2) = \log_e \sqrt{2}$. $	anh^{-1}\left(\frac{1}{3}ight) = \frac{1}{2} \log_e \left(\frac{1+1/3}{1-1/3}ight) = \frac{1}{2} \log_e \left(\frac{4/3}{2/3}ight) = \frac{1}{2} \log_e(2) = \log_e \sqrt{2}$. $\operatorname{cosech}^{-1}(-\sqrt{3}) = \log_e \left(\frac{1 - \sqrt{1+(-\sqrt{3})^2}}{-\sqrt{3}}ight) = \log_e \left(\frac{1 - \sqrt{4}}{-\sqrt{3}}ight) = \log_e \left(\frac{1-2}{-\sqrt{3}}ight) = \log_e \left(\frac{-1}{-\sqrt{3}}ight) = \log_e \left(\frac{1}{\sqrt{3}}ight) = -\log_e \sqrt{3}$. Step $2$: Combine the terms. $\operatorname{coth}^{-1}(3) + 	anh^{-1}\left(\frac{1}{3}ight) - \operatorname{cosech}^{-1}(-\sqrt{3}) = \log_e \sqrt{2} + \log_e \sqrt{2} - (-\log_e \sqrt{3}) = \log_e \sqrt{2} + \log_e \sqrt{2} + \log_e \sqrt{3} = \log_e (\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{3}) = \log_e (2\sqrt{3})$.

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

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