Question

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

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

Answer 1

Answer: (A)

It is given that, tan (P / 2) and tan (Q / 2) are the roots of
the quadratic equation $ax^2 + bx + c = 0 $
and $ \angle R = \pi / 2 $
$\therefore $ tan ( P / 2) + tan (Q/ 2) = - b / a
and tan (P / 2) tan (Q / 2) = c/a
Since, P + Q + R = $ 180^\circ$
$\Rightarrow P + Q = 90^\circ $
$\Rightarrow \frac{ P + Q }{ 2} = 45^\circ $
$\Rightarrow tan \, \bigg( \frac{ P + Q }{ 2} \bigg) = tan \, 45 ^\circ $
$\Rightarrow \frac{ tan \, (P / 2 ) + tan \, (Q/2)}{ 1 - tan \, (P / 2) \, tan \, (Q/ 2)} = 1 \Rightarrow \frac{ - b/a}{ 1 - c/a } = 1 $
$\Rightarrow \frac{ - b/a}{\frac{a - c}{a}} = 1 \Rightarrow \frac{ - b}{ a - c} = 1 $
$\Rightarrow -b = a - c \Rightarrow a + b = c $