If x = operatorname{sech}^{-1} frac{1}{2} + t | vedclass

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(C) Given that,$x = \operatorname{sech}^{-1} \frac{1}{2} + 	anh^{-1} \frac{1}{2}$.Using the logarithmic forms: $\operatorname{sech}^{-1} z = \ln \lef

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$\frac{4 \sqrt{3} + 3}{3}$

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(C) Given that,$x = \operatorname{sech}^{-1} \frac{1}{2} + 	anh^{-1} \frac{1}{2}$.Using the logarithmic forms: $\operatorname{sech}^{-1} z = \ln \left( \frac{1 + \sqrt{1 - z^2}}{z} ight)$ and $	anh^{-1} z = \frac{1}{2} \ln \left( \frac{1 + z}{1 - z} ight)$.For $z = \frac{1}{2}$,$\operatorname{sech}^{-1} \frac{1}{2} = \ln \left( \frac{1 + \sqrt{1 - 1/4}}{1/2} ight) = \ln \left( 2(1 + \frac{\sqrt{3}}{2}) ight) = \ln (2 + \sqrt{3})$.And $	anh^{-1} \frac{1}{2} = \frac{1}{2} \ln \left( \frac{1 + 1/2}{1 - 1/2} ight) = \frac{1}{2} \ln (3) = \ln \sqrt{3}$.Thus,$x = \ln (2 + \sqrt{3}) + \ln \sqrt{3} = \ln (\sqrt{3}(2 + \sqrt{3})) = \ln (2\sqrt{3} + 3)$.Now,$\cosh x = \frac{e^x + e^{-x}}{2}$.Since $e^x = 2\sqrt{3} + 3$,then $e^{-x} = \frac{1}{2\sqrt{3} + 3} = \frac{2\sqrt{3} - 3}{(2\sqrt{3})^2 - 3^2} = \frac{2\sqrt{3} - 3}{12 - 9} = \frac{2\sqrt{3} - 3}{3}$.Therefore,$\cosh x = \frac{(2\sqrt{3} + 3) + \frac{2\sqrt{3} - 3}{3}}{2} = \frac{6\sqrt{3} + 9 + 2\sqrt{3} - 3}{6} = \frac{8\sqrt{3} + 6}{6} = \frac{4\sqrt{3} + 3}{3}$.

If $x = \operatorname{sech}^{-1} \frac{1}{2} + \tanh^{-1} \frac{1}{2}$,then $\cosh x =$

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