2 coth^{-1}(4) + text{sech}^{-1}left(frac{3}{ | vedclass

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(A) We know that $\coth^{-1}(x) = \frac{1}{2} \log \left(\frac{x+1}{x-1}\right)$ for $|x| > 1$. Thus,$2 \coth^{-1}(4) = 2 \cdot \frac{1}{2} \log \left

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$\log 5$

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(A) We know that $\coth^{-1}(x) = \frac{1}{2} \log \left(\frac{x+1}{x-1}ight)$ for $|x| > 1$. Thus,$2 \coth^{-1}(4) = 2 \cdot \frac{1}{2} \log \left(\frac{4+1}{4-1}ight) = \log \left(\frac{5}{3}ight)$. We also know that $	ext{sech}^{-1}(x) = \log \left(\frac{1 + \sqrt{1-x^2}}{x}ight)$ for $0 Substituting $x = \frac{3}{5}$,we get $	ext{sech}^{-1}\left(\frac{3}{5}ight) = \log \left(\frac{1 + \sqrt{1 - (9/25)}}{3/5}ight) = \log \left(\frac{1 + \sqrt{16/25}}{3/5}ight) = \log \left(\frac{1 + 4/5}{3/5}ight) = \log \left(\frac{9/5}{3/5}ight) = \log 3$. Therefore,$2 \coth^{-1}(4) + 	ext{sech}^{-1}\left(\frac{3}{5}ight) = \log \left(\frac{5}{3}ight) + \log 3 = \log \left(\frac{5}{3} 	imes 3ight) = \log 5$.

$2 \coth^{-1}(4) + \text{sech}^{-1}\left(\frac{3}{5}\right) = $

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