Question
In a triangle $$PQR,$$ $$\angle R = \frac{\pi }{2}.$$ If $$\tan \left( {\frac{P}{2}} \right)$$ and $$\tan \left( {\frac{Q}{2}} \right)$$ are the roots of the equation $$a{x^2} + bx + c = 0\left( {a \ne 0} \right)$$ then.
A.
$$a + b = c$$
B.
$$b + c = a$$
C.
$$a + c = b$$
D.
$$b = c$$
Answer :
$$a + b = c$$
Solution :
Given that in $$\Delta PQR,\angle R = \frac{\pi }{2}$$
$$\eqalign{
& \Rightarrow \,\,\angle P + \angle Q = \frac{\pi }{2} \cr
& \Rightarrow \,\,\frac{{\angle P}}{2} + \frac{{\angle Q}}{2} = \frac{\pi }{4} \cr} $$
Also $$\tan \frac{P}{2}{\text{ and tan}}\frac{Q}{2}$$ are roots of the equation
$$\eqalign{
& a{x^2} + bx + c = 0\left( {a \ne 0} \right) \cr
& \therefore \,\,\tan \frac{P}{2} + \tan \frac{Q}{2} = - \frac{b}{a};\tan \frac{P}{2}\tan \frac{Q}{2} = \frac{c}{a} \cr} $$
Now consider, $$\tan \left( {\frac{{P + Q}}{2}} \right) = \frac{{\tan \frac{P}{2} + \tan \frac{Q}{2}}}{{1 - \tan \frac{P}{2}\tan \frac{Q}{2}}}$$
$$\eqalign{
& \Rightarrow \,\,\tan \frac{\pi }{4} = \frac{{ - \frac{b}{a}}}{{1 - \frac{c}{a}}} \cr
& \Rightarrow \,\,1 - \frac{c}{a} = - \frac{b}{a} \cr
& \Rightarrow \,\,a - c = - b \cr
& \Rightarrow \,\,a + b = c \cr} $$