sec h^{-1}left(frac{1}{2}right)-operatorname{ | vedclass

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(C) We use the logarithmic forms of inverse hyperbolic functions:$\sec h^{-1} x = \log _e\left(\frac{1+\sqrt{1-x^2}}{x}\right)$ and $\operatorname{cos

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

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(C) We use the logarithmic forms of inverse hyperbolic functions:$\sec h^{-1} x = \log _e\left(\frac{1+\sqrt{1-x^2}}{x}\right)$ and $\operatorname{cosec} h^{-1} x = \log _e\left(\frac{1+\sqrt{1+x^2}}{x}\right)$.For $\sec h^{-1}\left(\frac{1}{2}\right)$:$\sec h^{-1}\left(\frac{1}{2}\right) = \log _e\left(\frac{1+\sqrt{1-(1/2)^2}}{1/2}\right) = \log _e\left(\frac{1+\sqrt{3/4}}{1/2}\right) = \log _e\left(\frac{1+\sqrt{3}/2}{1/2}\right) = \log _e(2+\sqrt{3})$.For $\operatorname{cosec} h^{-1}\left(\frac{3}{4}\right)$:$\operatorname{cosec} h^{-1}\left(\frac{3}{4}\right) = \log _e\left(\frac{1+\sqrt{1+(3/4)^2}}{3/4}\right) = \log _e\left(\frac{1+\sqrt{25/16}}{3/4}\right) = \log _e\left(\frac{1+5/4}{3/4}\right) = \log _e\left(\frac{9/4}{3/4}\right) = \log _e(3)$.Subtracting the two values:$\log _e(2+\sqrt{3}) - \log _e(3) = \log _e\left(\frac{2+\sqrt{3}}{3}\right)$.

$\sec h^{-1}\left(\frac{1}{2}\right)-\operatorname{cosec} h^{-1}\left(\frac{3}{4}\right)$ is equal to

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