Question

In a $△PQR,\angle R=\frac{\pi }{2},iftan\left(\frac{P}{2}\right)$ and {\,} tan $\left(\frac{Q}{2}\right)$ are the
roots of the equation $a{x}^2+bx+c=0(a\ne 0)$, 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 $a{x}^2+bx+c=0$
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 = $180^{\circ }$
$\Rightarrow P+Q=90^{\circ }$
$\Rightarrow \frac{P+Q}{2}=45^{\circ }$
$\Rightarrow tan\left(\frac{P+Q}{2}\right)=tan45^{\circ }$
$\Rightarrow \frac{tan(P/2)+tan(Q/2)}{1-tan(P/2)tan(Q/2)}=1\Rightarrow \frac{-b/a}{1-c/a}=1$
$\Rightarrow \frac{-b/a}{\frac{a-c}{a}}=1\Rightarrow \frac{-b}{a-c}=1$
$\Rightarrow -b=a-c\Rightarrow a+b=c$