1. For the inverse cosecant function $\csc^{-1}(t)$ to be defined, its argument must satisfy $|t|\ge 1.$ So we need $\left|\frac{1+x}{x}\right|\ge 1, \qquad x\ne 0.$

2. Rewrite the argument: $\frac{1+x}{x}=1+\frac{1}{x}.$ Thus we solve $\left|\frac{1+x}{x}\right|\ge 1.$

3. Square both sides (valid since both sides are nonnegative under modulus): $\left(\frac{1+x}{x}\right)^2\ge 1$ $\frac{(1+x)^2}{x^2}\ge 1.$ Since $x^2>0$ for $x\ne 0$, multiply through by $x^2$: $(1+x)^2\ge x^2.$

4. Expand: $x^2+2x+1\ge x^2$ $2x+1\ge 0$ $x\ge -\frac{1}{2}.$

5. Also, $x\ne 0$ because the expression $\dfrac{1+x}{x}$ is undefined at $x=0$.

Hence the domain is $\left[-\frac12,\infty\right)\setminus\{0\}.$

6. Match with options:

• A: incorrect
• B: incorrect
• C: incorrect because it excludes $x=-\frac12$, but at $x=-\frac12$, $\frac{1+x}{x}=\frac{1/2}{-1/2}=-1,$ and $\csc^{-1}(-1)$ is defined.
• D: correct

Therefore, the correct option is D.