With usual notations in $\triangle \mathrm{ABC}$, if $\angle \mathrm{B}=\frac{\pi}{2}$, and $\tan \frac{\mathrm{A}}{2}, \tan \frac{\mathrm{C}}{2}$ are roots of equation $\mathrm{p} x^2+\mathrm{q} x+\mathrm{r}=0$, $\mathrm{p} \neq 0$, then

In the right triangle $\triangle ABC$ with $\angle B = \frac{\pi}{2}$, the sum of the remaining angles gives $\angle A + \angle C = \frac{\pi}{2}$, so $\frac{A}{2} + \frac{C}{2} = \frac{\pi}{4}$.

Taking tangent on both sides and applying the tangent addition formula yields $\frac{\tan\frac{A}{2} + \tan\frac{C}{2}}{1 - \tan\frac{A}{2} \tan\frac{C}{2}} = 1$, which simplifies to $\tan\frac{A}{2} + \tan\frac{C}{2} + \tan\frac{A}{2} \tan\frac{C}{2} = 1$.

Given that $\tan\frac{A}{2}$ and $\tan\frac{C}{2}$ are roots of $px^2 + qx + r = 0$, by Vieta's formulas, their sum is $-\frac{q}{p}$ and product is $\frac{r}{p}$, so $-\frac{q}{p} + \frac{r}{p} = 1$.

Multiplying through by $p$ and rearranging gives the relation $p + q = r$.

Answer: A