Question

In a triangle $$PQR,\ if\ \angle R=\dfrac{\pi}{2}$$ and  $$\displaystyle \tan(\frac{P}{2})$$ , $$\displaystyle \tan(\frac{Q}{2})$$ are the roots of the equation $$ax^{2}+bx+c=0(a\neq 0)$$ , then

A
a+b=c
B
b+c=a
C
c+a=b
D
b=c

Solution

The correct option is A $$a+b=c$$Given In $$\triangle PQR\ \angle R=90^\circ,\tan\dfrac{ P}{2} , \tan \dfrac{Q}{2}$$ are roots of the equation $$ax^2+bx+c=0(a\not =0)$$$$P+Q+R=\pi\Rightarrow P+Q+\dfrac{\pi}{2}=\pi$$$$P+Q = \dfrac {\pi}{2}$$ ,  sum of roots$$= \tan\dfrac{ P}{2} + \tan \dfrac{Q}{2} = \dfrac{-b}{a}$$$$\dfrac{P+Q}{2} = \cfrac{\pi}{4}$$,   Products of roots$$= \tan \dfrac{P}{2} . \tan\dfrac{ Q}{2} = \dfrac{c}{a}$$ $$\tan \big(\dfrac{P+Q}{2}\big) = 1$$$$\Rightarrow \cfrac{\tan P/2 + \tan Q/2}{1-\tan P/2 \tan Q/2} = 1$$$$\Rightarrow \cfrac{\dfrac{-b}{a}}{1-\dfrac{c}{a}} = 1$$$$\Rightarrow \cfrac{b}{c-a} = 1$$$$\Rightarrow a+b = c$$Maths

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