$\begin{array}{ll}
& \text { In } \triangle \mathrm{ABC}, \\
& \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ} \\
\therefore \quad & \angle \mathrm{A}+\frac{\pi}{2}+\angle \mathrm{B}=180^{\circ} \\
\therefore \quad & \angle \mathrm{A}+\angle \mathrm{B}=\frac{\pi}{2} \\
\therefore \quad & \frac{\angle \mathrm{A}}{2}+\frac{\angle \mathrm{B}}{2}=\frac{\pi}{4}
\end{array}$
$\tan \left(\frac{A}{2}\right)$ and $\tan \left(\frac{B}{2}\right)$ are roots of equation $\mathrm{a}_1 x^2+\mathrm{b}_1 x+\mathrm{c}_1=0 ... [Given]$
$\begin{aligned} & \therefore \quad \text { Sum of roots }=\frac{-b_1}{a_1} \\ & \tan \left(\frac{A}{2}\right)+\tan \left(\frac{B}{2}\right)=\frac{-b_1}{a_1}\end{aligned}$
Also, $\tan \left(\frac{\mathrm{A}}{2}\right) \cdot \tan \left(\frac{\mathrm{B}}{2}\right)=\frac{\mathrm{c}_1}{\mathrm{a}_1}$
Using $\tan \left(\frac{\mathrm{A}}{2}+\frac{\mathrm{B}}{2}\right)=\frac{\tan \frac{\mathrm{A}}{2}+\tan \frac{\mathrm{B}}{2}}{1-\tan \frac{\mathrm{A}}{2} \tan \frac{\mathrm{B}}{2}}$, we get
$\begin{aligned}
& \tan \left(\frac{\pi}{4}\right)=\frac{\frac{-b_1}{a_1}}{1-\frac{c_1}{a_1}} \\
& 1=\frac{-b_1}{a_1-c_1} \\
& a_1-c_1=-b_1 \\
& a_1+b_1=c_1
\end{aligned}$