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

$=\frac{\cosh \mathrm{\alpha }}{1-\tanh \mathrm{\alpha }}+\frac{\sinh \mathrm{\alpha }}{1-\coth \mathrm{\alpha }}$

$=\frac{\mathrm{cosh\alpha }}{1-{\displaystyle \frac{\mathrm{sinh\alpha }}{\mathrm{cosh\alpha }}}}+\frac{\sinh \mathrm{\alpha }}{1-{\displaystyle \frac{\mathrm{cosh\alpha }}{\sin h\alpha }}}$

$=\frac{\cos h^2\alpha }{\cos h\alpha -\sin h\alpha }+\frac{\sin h^2\alpha }{\sin h\alpha -\cos h\alpha }$

$=\frac{\cos h^2\alpha -\sin h^2\alpha }{\cos h\alpha -\sin h\alpha }$

$=\sin h\alpha +\cos h\alpha$

$=e^x$

$=2+\sqrt{3}$