If $\cot 60^{\circ }+\csc 60^{\circ }=x$, then the value of $x$ is:

We know the standard trigonometric values:
$$\cot 60^{\circ }=\frac{1}{\sqrt{3}}$$
$$\csc 60^{\circ }=\frac{2}{\sqrt{3}}$$
Substitute these values into the given equation:
$$x=\cot 60^{\circ }+\csc 60^{\circ }$$
$$x=\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}$$
$$x=\frac{3}{\sqrt{3}}$$
$$x=\sqrt{3}$$
(Note: Option D contains a small typo in the raw question text where the square root symbol was omitted/misprinted as 3 3, but the mathematically accurate option value is $\sqrt{3}$).