In a \(\triangle \mathrm{PQR}, \mathrm{m} \angle \mathrm{R}=\frac{\pi}{2}\). If \(\tan \left(\frac{\mathrm{P}}{2}\right)\) and \(\tan \left(\frac{\mathrm{Q}}{2}\right)\) are the roots of the equation \(a x^2+b x+c=0(a \neq 0)\), then

\(\text {In } \triangle \mathrm{PQR} \)
\( \angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}=180^{\circ} \)
\( \therefore \angle \mathrm{P}+\angle \mathrm{Q}+\frac{\pi}{2}=180^{\circ} \)
\( \therefore \angle \mathrm{P}+\angle \mathrm{Q}=\frac{\pi}{2} \)
\( \therefore \frac{\angle \mathrm{P}}{2}+\frac{\angle \mathrm{Q}}{2}=\frac{\pi}{4}\)
\(\tan \left(\frac{\mathrm{P}}{2}\right)\) and \(\tan \left(\frac{\mathrm{Q}}{2}\right)\) are roots of the equation
\(a x^2+b x+c=0\)...[Given]
\(\therefore\) Sum of roots \(=\frac{-b}{a}\)
\(\tan \left(\frac{\mathrm{P}}{2}\right)+\tan \left(\frac{\mathrm{Q}}{2}\right)=\frac{-\mathrm{b}}{\mathrm{a}}\)
Also, \(\tan \left(\frac{\mathrm{P}}{2}\right) \cdot \tan \left(\frac{\mathrm{Q}}{2}\right)=\frac{\mathrm{c}}{\mathrm{a}}\)
Using, \(\tan \left(\frac{P}{2}+\frac{Q}{2}\right)=\frac{\tan \frac{P}{2}+\tan \frac{Q}{2}}{1-\tan \frac{P}{2} \tan \frac{Q}{2}}\), we get
\(\Rightarrow \tan \left(\frac{\pi}{4}\right)=\frac{\frac{-b}{a}}{1-\frac{c}{a}} \)
\( \Rightarrow 1=\frac{-b}{a-c} \)
\( \Rightarrow \mathrm{a}-\mathrm{c}=-\mathrm{b} \)
\( \Rightarrow \mathrm{a}+\mathrm{b}=\mathrm{c}\)