Question

In a $△PQR$, if $\angle R=\frac{\pi }{2}$ 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. $b=a+c$
• B. $b=c$
• C. $c=a+b$
• D. $a=b+c$

Answer 1

The correct option is A $b=a+c$

$\tan \left(\frac{P+Q}{2}\right)=\frac{\tan \frac{P}{2}+\tan \frac{Q}{2}}{1-\tan \frac{P}{2}\tan \frac{Q}{2}}$
$\tan \frac{P}{2}+\tan \frac{Q}{2}=\frac{-b}{a}$
$\tan \frac{P}{2}\tan \frac{Q}{2}=\frac{c}{a}$

$\tan \left(\frac{\pi -R}{2}\right)=\frac{\frac{-b}{a}}{1-\frac{c}{a}}$

$\cot \frac{R}{2}=\frac{b}{c-a}\Rightarrow (c-a)\cot \frac{\pi }{4}=b$
$\Rightarrow c=a+b$