The domain of the function $$\operatorname{cosec}^{-1}\left(\frac{1+x}{x}\right)$$ is:

We have the composite expression $$\csc^{-1}\!\left(\dfrac{1+x}{x}\right)$$. For the inverse cosecant (written as $$\csc^{-1}(y)$$) to be defined, its argument $$y$$ must satisfy the basic condition derived from the cosecant function:

$$|\csc\theta| \;\ge\; 1 \quad\Longrightarrow\quad |\;y\;| \;\ge\; 1,$$

because cosecant values never lie in the open interval $$(-1,\,1)$$. Hence we require

$$\left|\dfrac{1+x}{x}\right| \;\ge\; 1.$$

There is also the obvious restriction $$x \neq 0$$, since the denominator in $$\dfrac{1+x}{x}$$ cannot be zero. Now we proceed to solve the inequality step by step.

First, recall the property of absolute value that

$$|A| \;\ge\; 1 \quad\Longleftrightarrow\quad A^2 \;\ge\; 1,$$

because squaring a non-negative number preserves the inequality. Setting $$A=\dfrac{1+x}{x}$$, we write

$$\left(\dfrac{1+x}{x}\right)^{\!2} \;\ge\; 1.$$

Expanding the square in the numerator and denominator, we obtain

$$\dfrac{(1+x)^2}{x^2} \;\ge\; 1.$$

To clear the denominator, multiply both sides by $$x^2$$. Since $$x^2$$ is always non-negative, the direction of the inequality remains unchanged:

$$(1+x)^2 \;\ge\; x^2.$$

Next, expand $$\bigl(1 + x\bigr)^2$$:

$$(1 + x)^2 \;=\; 1 + 2x + x^2.$$

Substituting this into the inequality gives

$$1 + 2x + x^2 \;\ge\; x^2.$$

Now subtract $$x^2$$ from both sides to simplify:

$$1 + 2x + x^2 - x^2 \;\ge\; 0 \quad\Longrightarrow\quad 1 + 2x \;\ge\; 0.$$

The $$x^2$$ terms cancel, leaving the linear inequality

$$2x + 1 \;\ge\; 0.$$

Solve for $$x$$ by first subtracting $$1$$:

$$2x \;\ge\; -1,$$

and then dividing by $$2$$ (which is positive, so the inequality direction is preserved):

$$x \;\ge\; -\dfrac{1}{2}.$$

Combining this result with the earlier restriction $$x \neq 0$$, the set of permissible $$x$$-values is

$$\left[-\dfrac{1}{2},\,\infty\right) \;-\;\{0\}.$$

Therefore, the domain of the function $$\csc^{-1}\!\left(\dfrac{1+x}{x}\right)$$ is exactly

$$\left[-\dfrac{1}{2},\,\infty\right) - \{0\}.$$

Hence, the correct answer is Option A.