In a triangle ABC with usual notations if, $\tan \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)=x \cot \frac{\mathrm{~A}}{2}$, then $x=$

Using Napier's Analogy, the trigonometric identity for $\tan \left(\frac{B-C}{2}\right)$ relates sides and angles as $\tan \left(\frac{B-C}{2}\right) = \frac{b-c}{b+c} \cot \frac{A}{2}$.

The given equation $\tan \left(\frac{B-C}{2}\right) = x \cot \frac{A}{2}$ matches this form.
Equating the expressions for $\tan \left(\frac{B-C}{2}\right)$ yields $x \cot \frac{A}{2} = \frac{b-c}{b+c} \cot \frac{A}{2}$.

Since $\cot \frac{A}{2} \neq 0$ for any triangle angle $A \in (0, \pi)$, division gives $x = \frac{b-c}{b+c}$.

Final answer: $\boxed{\text{C}}$