If operatorname{cosech} x = frac{4}{5},then c | vedclass

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(D) We know that $\operatorname{cosech} x = \frac{1}{\sinh x}$.Given $\operatorname{cosech} x = \frac{4}{5}$,therefore $\sinh x = \frac{5}{4}$.Using t

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$\sqrt{\frac{41}{16}}$

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(D) We know that $\operatorname{cosech} x = \frac{1}{\sinh x}$.Given $\operatorname{cosech} x = \frac{4}{5}$,therefore $\sinh x = \frac{5}{4}$.Using the identity $\cosh^2 x - \sinh^2 x = 1$,we have $\cosh^2 x = 1 + \sinh^2 x$.Substituting the value of $\sinh x$: $\cosh^2 x = 1 + (\frac{5}{4})^2 = 1 + \frac{25}{16} = \frac{16 + 25}{16} = \frac{41}{16}$.Since $\cosh x > 0$ for all real $x$,we take the positive square root: $\cosh x = \sqrt{\frac{41}{16}}$.

If $\operatorname{cosech} x = \frac{4}{5}$,then $\cosh x =$

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