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

Solution :
$$\tan \left( {\frac{P}{2}} \right),\tan \left( {\frac{Q}{2}} \right)$$    are the roots of $$a{x^2} + bx + c = 0$$
$$\eqalign{ & \tan \left( {\frac{P}{2}} \right) + \tan \left( {\frac{Q}{2}} \right) = - \frac{b}{a},\tan \left( {\frac{P}{2}} \right) \cdot \tan \left( {\frac{Q}{2}} \right) = \frac{c}{a} \cr & \frac{{\tan \left( {\frac{P}{2}} \right) + \tan \left( {\frac{Q}{2}} \right)}}{{1 - \tan \left( {\frac{P}{2}} \right)\tan \left( {\frac{Q}{2}} \right)}} = \tan \left( {\frac{P}{2} + \frac{Q}{2}} \right) = 1 \cr & \Rightarrow \,\,\frac{{ - \frac{b}{a}}}{{1 - \frac{c}{a}}} = 1 \cr & \Rightarrow \,\, - \frac{b}{a} = \frac{a}{a} - \frac{c}{a} \cr & \Rightarrow \,\, - b = a - c\,\,\,{\text{or }}c = a + b. \cr} $$