Question

In a $△PQR,\angle R=\frac{\pi }{2}.$ If $\tan \frac{P}{2}$ and $\tan \frac{Q}{2}$ are the roots of the equation $a{x}^2+bx+c=0$ then

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

Answer 1

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

$\tan \frac{P}{2}+\tan \frac{Q}{2}=-\frac{b}{a},\tan \frac{P}{2}.\tan \frac{Q}{2}=\frac{c}{a}$

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

Also given $\angle R={90}^0\therefore \tan \left(\frac{P+Q}{2}\right)=1$

$\Rightarrow \frac{b}{c-a}=1\Rightarrow a+b=c$