If cosh alpha + sinh alpha = e^3 and sinh x = | vedclass

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(A) Given that $\cosh \alpha + \sinh \alpha = e^3$. Using the definitions $\cosh \alpha = \frac{e^\alpha + e^{-\alpha}}{2}$ and $\sinh \alpha = \frac{

Vedclass · Accepted Answer

$\frac{\alpha}{\alpha+2}$

Vedclass · Answer

(A) Given that $\cosh \alpha + \sinh \alpha = e^3$. Using the definitions $\cosh \alpha = \frac{e^\alpha + e^{-\alpha}}{2}$ and $\sinh \alpha = \frac{e^\alpha - e^{-\alpha}}{2}$,we get: $\frac{e^\alpha + e^{-\alpha}}{2} + \frac{e^\alpha - e^{-\alpha}}{2} = e^3$ $e^\alpha = e^3 \Rightarrow \alpha = 3$. Now,substitute $\alpha = 3$ into the expression for $\sinh x$: $\sinh x = \frac{3}{3+1} = \frac{3}{4}$. We know that $\cosh^2 x - \sinh^2 x = 1$,so $\cosh x = \sqrt{1 + \sinh^2 x} = \sqrt{1 + (\frac{3}{4})^2} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$. Therefore,$	anh x = \frac{\sinh x}{\cosh x} = \frac{3/4}{5/4} = \frac{3}{5}$. Since $\alpha = 3$,we check the options: $\frac{\alpha}{\alpha+2} = \frac{3}{3+2} = \frac{3}{5}$. Thus,the correct option is $A$.

If $\cosh \alpha + \sinh \alpha = e^3$ and $\sinh x = \frac{\alpha}{\alpha+1}$,then $\tanh x =$

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