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

JEE Main · 2005 · 08 May 2005 · Offline · Properties of Triangle

In a triangle

PQR

∠ R = \frac{π}{2}

. If

tan (\frac{P}{2})

and

tan (\frac{Q}{2})

are the roots of

a x^{2} + b x + c = 0, a \neq = 0

then

Answer: (B)

Since

∠ R = \frac{π}{2}

, we have

P + Q = \frac{π}{2}

, so

\frac{P}{2} + \frac{Q}{2} = \frac{π}{4}

and

tan (\frac{P}{2} + \frac{Q}{2}) = 1

. Expanding,

\frac{tan \frac{P}{2} + tan \frac{Q}{2}}{1 - tan \frac{P}{2} tan \frac{Q}{2}} = 1

. As

tan \frac{P}{2}

and

tan \frac{Q}{2}

are the roots of

a x^{2} + b x + c = 0

, their sum is

- \frac{b}{a}

and their product is

\frac{c}{a}

. Substituting,

\frac{- b / a}{1 - c / a} = 1

, so

- b = a - c

, which gives

c = a + b

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option