Let $E_1=\left\{x \in \mathrm{R}: x \neq 1\right.$ and $\left.\frac{x}{x-1}>0\right\}$
and $E_2=\left\{x \in \mathrm{E}_1: \sin ^{-1}\left(\log _{\mathrm{e}}\left(\frac{x}{x-1}\right)\right)\right.$ is a real number $\}$
(Here, the inverse trigonometric function $\sin ^{-1} \mathrm{x}$ assumes values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.)
Let $\quad f: E_1 \rightarrow \mathrm{R}$ be the function defined by $f(x)=\log _e\left(\frac{x}{x-1}\right)$
and $\quad g: E_2 \rightarrow \mathrm{R}$ be the function defined by $g(x)=\sin ^{-1}\left(\log _e\left(\frac{x}{x-1}\right)\right)$. The correct option is :