If alpha = log_e(2+sqrt{3}),then frac{cosh al | vedclass

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(D) Let $I = \frac{\cosh \alpha}{1-	anh \alpha} + \frac{\sinh \alpha}{1-\coth \alpha}$.Substitute $	anh \alpha = \frac{\sinh \alpha}{\cosh \alpha}$

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$2+\sqrt{3}$

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(D) Let $I = \frac{\cosh \alpha}{1-	anh \alpha} + \frac{\sinh \alpha}{1-\coth \alpha}$.Substitute $	anh \alpha = \frac{\sinh \alpha}{\cosh \alpha}$ and $\coth \alpha = \frac{\cosh \alpha}{\sinh \alpha}$:$I = \frac{\cosh \alpha}{1 - \frac{\sinh \alpha}{\cosh \alpha}} + \frac{\sinh \alpha}{1 - \frac{\cosh \alpha}{\sinh \alpha}}$$I = \frac{\cosh^2 \alpha}{\cosh \alpha - \sinh \alpha} + \frac{\sinh^2 \alpha}{\sinh \alpha - \cosh \alpha}$$I = \frac{\cosh^2 \alpha - \sinh^2 \alpha}{\cosh \alpha - \sinh \alpha}$Since $\cosh^2 \alpha - \sinh^2 \alpha = 1$,we have $I = \frac{1}{\cosh \alpha - \sinh \alpha}$.Using the identity $\cosh \alpha - \sinh \alpha = e^{-\alpha}$,we get $I = \frac{1}{e^{-\alpha}} = e^{\alpha}$.Given $\alpha = \log_e(2+\sqrt{3})$,then $e^{\alpha} = 2+\sqrt{3}$.

If $\alpha = \log_e(2+\sqrt{3})$,then $\frac{\cosh \alpha}{1-\tanh \alpha} + \frac{\sinh \alpha}{1-\coth \alpha} = $

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