How to find the domain of
$$\csc^{-1}\left(\frac{x+1}{x}\right)$$
My attempt :
the cosecant function lies in $\,\Bbb R\setminus(-1,1)\;$ ,
so first checking the range of $\;\dfrac{x+1}{x}\;$ ,
we find that it covers all reals but $1$.
Thus the domain of the function must be $\,\Bbb R\setminus(-1,1)$.
But this is wrong.
Why is this so ?
Source : JEE Mains ,2021, august attempt
Since the domain of the function $\;\color{blue}{\csc^{-1}}$ is $\;\color{brown}{\Bbb R\setminus(-1,1)}\,,\;$ that is $\;\color{brown}{\big(\!-\!\infty,-1\big]\cup\big[1,+\infty\big)}\,,\;$ you should solve the following inequalities :
$\dfrac{x+1}x\leqslant-1\quad\lor\quad\dfrac{x+1}x\geqslant1\;.$
The solutions of the first inequality are :
$S_1=\left[-\dfrac12,0\right)\;$.
The solutions of the second inequality are :
$S_2=\big(0,+\infty\big)\;.$
Hence , the domain of the function $\;f(x)=\csc^{-1}\!\left(\dfrac{x+1}x\right)\;$ is
$\left[-\dfrac12,0\right)\cup\big(0,+\infty\big)\;.$