If 5 sinh x - cosh x = 5,then one of the valu | vedclass

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(C) Given $5 \sinh x - \cosh x = 5$. Rearranging,we get $5 \sinh x - 5 = \cosh x$,which implies $5(\sinh x - 1) = \cosh x$. Squaring both sides,we hav

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$\frac{-3}{5}$

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(C) Given $5 \sinh x - \cosh x = 5$. Rearranging,we get $5 \sinh x - 5 = \cosh x$,which implies $5(\sinh x - 1) = \cosh x$. Squaring both sides,we have $25(\sinh^2 x + 1 - 2 \sinh x) = \cosh^2 x$. Using the identity $\cosh^2 x - \sinh^2 x = 1$,we substitute $\cosh^2 x = 1 + \sinh^2 x$: $25 \sinh^2 x + 25 - 50 \sinh x = 1 + \sinh^2 x$. $24 \sinh^2 x - 50 \sinh x + 24 = 0$. Dividing by $2$,we get $12 \sinh^2 x - 25 \sinh x + 12 = 0$. Factoring the quadratic: $(3 \sinh x - 4)(4 \sinh x - 3) = 0$. Thus,$\sinh x = \frac{4}{3}$ or $\sinh x = \frac{3}{4}$. If $\sinh x = \frac{4}{3}$,then $\cosh x = 5(\frac{4}{3} - 1) = \frac{5}{3}$,so $	anh x = \frac{\sinh x}{\cosh x} = \frac{4/3}{5/3} = \frac{4}{5}$. If $\sinh x = \frac{3}{4}$,then $\cosh x = 5(\frac{3}{4} - 1) = -\frac{5}{4}$,so $	anh x = \frac{\sinh x}{\cosh x} = \frac{3/4}{-5/4} = -\frac{3}{5}$. Therefore,one of the values is $-\frac{3}{5}$.

If $5 \sinh x - \cosh x = 5$,then one of the values of $\tanh x$ is

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