operatorname{sech}^{-1}left(frac{3}{5}right)- | vedclass

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Question

(C) We need to evaluate $\operatorname{sech}^{-1}\left(\frac{3}{5}ight)-	anh ^{-1}\left(\frac{3}{5}ight)$.First, we use the identity $\operatorna

Vedclass · Accepted Answer

$\log _e\left(\frac{3}{2}\right)$

Vedclass · Answer

(C) We need to evaluate $\operatorname{sech}^{-1}\left(\frac{3}{5}ight)-	anh ^{-1}\left(\frac{3}{5}ight)$.First, we use the identity $\operatorname{sech}^{-1}(x) = \cosh^{-1}\left(\frac{1}{x}ight)$.So, $\operatorname{sech}^{-1}\left(\frac{3}{5}ight) = \cosh^{-1}\left(\frac{5}{3}ight)$.Using the logarithmic form $\cosh^{-1}(x) = \log_e(x + \sqrt{x^2 - 1})$:$\cosh^{-1}\left(\frac{5}{3}ight) = \log_e\left(\frac{5}{3} + \sqrt{\frac{25}{9} - 1}ight) = \log_e\left(\frac{5}{3} + \sqrt{\frac{16}{9}}ight) = \log_e\left(\frac{5}{3} + \frac{4}{3}ight) = \log_e(3)$.Next, we use the identity $	anh^{-1}(x) = \frac{1}{2} \log_e\left(\frac{1+x}{1-x}ight)$.So, $	anh^{-1}\left(\frac{3}{5}ight) = \frac{1}{2} \log_e\left(\frac{1 + 3/5}{1 - 3/5}ight) = \frac{1}{2} \log_e\left(\frac{8/5}{2/5}ight) = \frac{1}{2} \log_e(4) = \frac{1}{2} \log_e(2^2) = \log_e(2)$.Finally, subtracting the two values:$\operatorname{sech}^{-1}\left(\frac{3}{5}ight) - 	anh^{-1}\left(\frac{3}{5}ight) = \log_e(3) - \log_e(2) = \log_e\left(\frac{3}{2}ight)$.

$\operatorname{sech}^{-1}\left(\frac{3}{5}\right)-\tanh ^{-1}\left(\frac{3}{5}\right)=$

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