In a triangle $$PQR,\;\;\angle R = {\pi \over 2}.\,\,If\,\,\tan \,\left( {{P \over 2}} \right)$$ and $$ \tan \left( {{Q \over 2}} \right)$$ are the roots of $$a{x^2} + bx + c = 0,\,\,a \ne 0$$ then
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Explanation $$\angle $$R = 90o $$ \therefore $$ $$\angle $$P + $$\angle $$Q = 90o
$$ \Rightarrow $$ $${P \over 2} + {Q \over 2} = {{90} \over 2} = 45$$o
$$\tan \left( {{P \over 2}} \right),\tan \left( {{Q \over 2}} \right)$$ are the roots of $$a{x^2} + bx + c = 0$$
$$ \therefore $$ $$\tan \left( {{P \over 2}} \right) + \tan \left( {{Q \over 2}} \right) = - {b \over a},\,\,$$
and $$\tan \left( {{P \over 2}} \right).\tan \left( {{Q \over 2}} \right) = {c \over a}$$
$${{\tan \left( {{P \over 2}} \right) + \tan \left( {{Q \over 2}} \right)} \over {1 - \tan \left( {{P \over 2}} \right)\tan \left( {{Q \over 2}} \right)}}$$
$$ = \tan \left( {{P \over 2} + {Q \over 2}} \right) $$= tan 45o = 1
$$ \Rightarrow {{ - {b \over a}} \over {1 - {c \over a}}} = 1$$
$$ \Rightarrow - {b \over a} = {a \over a} - {c \over a}$$
$$ \Rightarrow - b = a - c$$ or $$c = a + b.$$