In a triangle $PQR$, $\angle R=\frac{\pi }{2}$. If $\tan \frac{P}{2}$ and $\tan \frac{Q}{2}$ are the roots of $a{x}^2+bx+c=0,a\ne 0$, then:

In $△PQR$, $P+Q+R=\pi$. Given $R=\pi /2$, then $P+Q=\pi /2$.Thus, $\frac{P}{2}+\frac{Q}{2}=\frac{\pi }{4}$.Taking tangent on both sides:
$$\tan \left(\frac{P}{2}+\frac{Q}{2}\right)=\tan \frac{\pi }{4}$$
$$\frac{\tan \frac{P}{2}+\tan \frac{Q}{2}}{1-\tan \frac{P}{2}\tan \frac{Q}{2}}=1$$
From the quadratic equation $a{x}^2+bx+c=0$, the sum of roots is $-b/a$ and the product of roots is $c/a$.
$$\frac{-b/a}{1-c/a}=1$$
$$\frac{-b/a}{(a-c)/a}=1$$
$$-b=a-c⟹c=a+b$$