If $\cot \left(\frac{A}{2}\right)=x$, then $x$ is equal to:

We know the trigonometric half-angle identity for cotangent:
$$\cot \left(\frac{A}{2}\right)=\frac{1+\cos A}{\sin A}$$
Let's simplify this by dividing each term in the numerator by $\sin A$:
$$\cot \left(\frac{A}{2}\right)=\frac{1}{\sin A}+\frac{1}{\tan A}=\csc A+\cot A$$
Using the algebraic identity $a^2-b^2=1$ for cosecant and cotangent:
$${\csc }^{2}A-{\cot }^{2}A=1$$
$$(\csc A-\cot A)(\csc A+\cot A)=1$$
$$\csc A+\cot A=\frac{1}{\csc A-\cot A}$$
Therefore:
$$x=\frac{1}{\csc A-\cot A}$$