In a ${\Delta }^{le}PQR,\overline{)R}=\frac{\pi }{2}$$\text{ if }\tan \left(\frac{P}{2}\right)\text{ and }\tan \left(\frac{Q}{2}\right)\text{ are the roots of the equation }a{x}^2+bx+c=0(a\ne 0)$

then 

$\mathrm{P}+\mathrm{Q}={90}^0 \frac{\mathrm{P}}{2}+\frac{\mathrm{Q}}{2}={45}^{0 }\tan \frac{\mathrm{P}}{2}+\tan \frac{\mathrm{Q}}{2}=-\frac{\mathrm{b}}{\mathrm{a}} \tan \frac{\mathrm{P}}{2}.\tan \frac{\mathrm{Q}}{2}=\frac{\mathrm{c}}{\mathrm{q}}$
  $$\begin{gathered} \tan \left(\frac{\mathrm{P}}{2}+\frac{\mathrm{Q}}{2}\right)=1 \\ \frac{\tan {\displaystyle \frac{P}{2}}+\tan {\displaystyle \frac{Q}{2}}}{1-\tan {\displaystyle \frac{P}{2}}\tan {\displaystyle \frac{Q}{2}}}=1 \\ \frac{-{\displaystyle \frac{b}{a}}}{1-{\displaystyle \frac{c}{a}}}=1\Rightarrow -\frac{b}{a-c}=1 \\ \Rightarrow -b=a-c\Rightarrow c=a+b \end{gathered}$$