If sinh x = -frac{4}{3},then sinh 2x + cosh 2 | vedclass

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(D) Given $\sinh x = -\frac{4}{3}$.We know the identity $\cosh^2 x - \sinh^2 x = 1$,which implies $\cosh^2 x = 1 + \sinh^2 x$.$\cosh^2 x = 1 + (-\frac

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$9$

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(D) Given $\sinh x = -\frac{4}{3}$.We know the identity $\cosh^2 x - \sinh^2 x = 1$,which implies $\cosh^2 x = 1 + \sinh^2 x$.$\cosh^2 x = 1 + (-\frac{4}{3})^2 = 1 + \frac{16}{9} = \frac{25}{9}$.Since $\cosh x \geq 1$ for all real $x$,we take the positive root: $\cosh x = \frac{5}{3}$.Now,$\sinh 2x + \cosh 2x = (2 \sinh x \cosh x) + (\cosh^2 x + \sinh^2 x)$.Substituting the values: $2(-\frac{4}{3})(\frac{5}{3}) + (\frac{25}{9} + \frac{16}{9})$.$= -\frac{40}{9} + \frac{41}{9} = \frac{1}{9}$.

If $\sinh x = -\frac{4}{3}$,then $\sinh 2x + \cosh 2x = $

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