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 $a{x}^2+bx+c=0,a\ne 0$ then

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

Answer 1

The correct option is B c = a + b

$tan\left(\frac{P}{2}\right),tan\left(\frac{Q}{2}\right)$ are the roots of $a{x}^2+bx+c=0$
$tan\left(\frac{P}{2}\right)+tan\left(\frac{Q}{2}\right)=-\frac{b}{a},tan\left(\frac{P}{2}\right).tan\left(\frac{Q}{2}\right)=\frac{c}{a}\frac{tan\left(\frac{P}{2}\right)+tan\left(\frac{Q}{2}\right)}{1-tan\left(\frac{P}{2}\right)tan\left(\frac{Q}{2}\right)}=tan\left(\frac{P+Q}{2}\right)=1\Rightarrow \frac{-\frac{b}{a}}{1-\frac{c}{a}}=1\Rightarrow -\frac{b}{a}=\frac{a}{a}-\frac{c}{a}\Rightarrow -b=a-corc=a+b.$