If sinh x = frac{sqrt{21}}{2},then cosh 2x + | vedclass

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(C) Given $\sinh x = \frac{\sqrt{21}}{2}$.We know that $\cosh^2 x - \sinh^2 x = 1$,so $\cosh x = \sqrt{1 + \sinh^2 x} = \sqrt{1 + \frac{21}{4}} = \sqr

Vedclass · Accepted Answer

$\frac{23 + 5\sqrt{21}}{2}$

Vedclass · Answer

(C) Given $\sinh x = \frac{\sqrt{21}}{2}$.We know that $\cosh^2 x - \sinh^2 x = 1$,so $\cosh x = \sqrt{1 + \sinh^2 x} = \sqrt{1 + \frac{21}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$.Using the identities $\sinh 2x = 2 \sinh x \cosh x$ and $\cosh 2x = \cosh^2 x + \sinh^2 x$:$\sinh 2x = 2 	imes \frac{\sqrt{21}}{2} 	imes \frac{5}{2} = \frac{5\sqrt{21}}{2}$.$\cosh 2x = (\frac{5}{2})^2 + (\frac{\sqrt{21}}{2})^2 = \frac{25}{4} + \frac{21}{4} = \frac{46}{4} = \frac{23}{2}$.Therefore,$\cosh 2x + \sinh 2x = \frac{23}{2} + \frac{5\sqrt{21}}{2} = \frac{23 + 5\sqrt{21}}{2}$.

If $\sinh x = \frac{\sqrt{21}}{2}$,then $\cosh 2x + \sinh 2x = $

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