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{co

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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 $\operatorname{cosec} h^{-1} x = \log _e\left(\frac{1+\sqrt{1+x^2}}{x}\right)$. Substituting $x = \frac{1}{2}$ in $\sec h^{-1} x$: $\sec h^{-1}\left(\frac{1}{2}\right) = \log _e\left(\frac{1+\sqrt{1-\frac{1}{4}}}{\frac{1}{2}}\right) = \log _e\left(\frac{1+\frac{\sqrt{3}}{2}}{\frac{1}{2}}\right) = \log _e(2+\sqrt{3})$. Substituting $x = \frac{3}{4}$ in $\operatorname{cosec} h^{-1} x$: $\operatorname{cosec} h^{-1}\left(\frac{3}{4}\right) = \log _e\left(\frac{1+\sqrt{1+\frac{9}{16}}}{\frac{3}{4}}\right) = \log _e\left(\frac{1+\frac{5}{4}}{\frac{3}{4}}\right) = \log _e\left(\frac{\frac{9}{4}}{\frac{3}{4}}\right) = \log _e(3)$. Therefore,$\sec h^{-1}\left(\frac{1}{2}\right) - \operatorname{cosec} h^{-1}\left(\frac{3}{4}\right) = \log _e(2+\sqrt{3}) - \log _e(3) = \log _e\left(\frac{2+\sqrt{3}}{3}\right)$.

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

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