<s>[INST] Find the domain of $\sec^{-1}(\frac{2+x}{1-x})^{\frac{1}{x}}$ Find the domain of $\sec^{-1}(\frac{2+x}{1-x})^{\frac{1}{x}}$

Since the domain of definition of $\sec^{-1} x$ is $x\leq-1$ or $x\geq 1$.
So $(\frac{2+x}{1-x})^{\frac{1}{x}}\leq -1$ or $(\frac{2+x}{1-x})^{\frac{1}{x}}\geq 1$
The domain of $(\frac{2+x}{1-x})^{\frac{1}{x}}$ is known to be $[-2,0)\cup(0,1)$.
I am stuck here.Please help.
 [/INST] $\Longrightarrow$The Range of $(\frac{2+x}{1-x})^{\frac{1}{x}}$ =
[0,$\infty$)
The domain of Sec$^{-1}x$ = $\mathbb{R}$- (-1,1) 
$\Longrightarrow$1$\leq$$\left(\frac{2+x}{1-x}\right)^{\frac{1}{x}}$$\Longrightarrow$
x$\geq$ -$\frac{1}{2}$$\Longrightarrow$ the domain of $\sec^{-1}(\frac{2+x}{1-x})^{\frac{1}{x}}$
is [$\frac{-1}{2}$,1) -$\left\{ 0\right\} $
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