In \(\triangle \mathrm{ABC}\), with tustal notations, \(\mathrm{m} \angle \mathrm{C}=\frac{\pi}{2}\), if \(\tan \left(\frac{\mathrm{A}}{2}\right)\) and \(\tan \left(\frac{\mathrm{B}}{2}\right)\) are the roots of the equation. \(\mathrm{a}_1 x^2+\mathrm{b}_1 x+\mathrm{c}_1=0\left(\mathrm{a}_1 \neq 0\right)\), then

\(\text {In } \triangle \mathrm{ABC}, \)
\( \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ} \)
\( \therefore \angle \mathrm{A}+\frac{\pi}{2}+\angle \mathrm{B}=180^{\circ} \)
\( \therefore \angle \mathrm{A}+\angle \mathrm{B}=\frac{\pi}{2} \)
\( \therefore \frac{\angle \mathrm{A}}{2}+\frac{\angle \mathrm{B}}{2}=\frac{\pi}{4}\)
\(\tan \left(\frac{A}{2}\right)\) and \(\tan \left(\frac{B}{2}\right)\) are roots of equation \(\mathrm{a}_1 x^2+\mathrm{b}_1 x+\mathrm{c}_1=0 ... [Given]\)
\(\therefore \text {Sum of roots }=\frac{-b_1}{a_1} \)
\( \tan \left(\frac{A}{2}\right)+\tan \left(\frac{B}{2}\right)=\frac{-b_1}{a_1}\)
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
\(\tan \left(\frac{\pi}{4}\right)=\frac{\frac{-b_1}{a_1}}{1-\frac{c_1}{a_1}} \)
\( 1=\frac{-b_1}{a_1-c_1} \)
\( a_1-c_1=-b_1 \)
\( a_1+b_1=c_1\)