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

Answer 1

Answer: (C)

If $\alpha$ and $\beta$ are the roots of the equation $ax+hx+c=0$, then $\alpha +\beta =\frac{-b}{a}$ and $\alpha \beta =\frac{c}{a}.$
Since, $\tan \left(\frac{P}{2}\right)$ and $\tan \left(\frac{Q}{2}\right)$ are roots of equation $a{x}^2+bx+c=0.$
$\therefore \tan \frac{P}{2}+\tan \frac{Q}{2}=-\frac{b}{a}$
and $\tan \frac{P}{2}\tan \frac{Q}{2}=\frac{c}{a}$
Also, $\frac{P}{2}+\frac{Q}{2}+\frac{R}{2}=\frac{\pi }{2}$
(As $P,Q,R$ are angles of a triangle)
$\Rightarrow \frac{P+Q}{2}=\frac{\pi }{2}-\frac{R}{2}\Rightarrow \frac{P+Q}{2}=\frac{\pi }{4}$
Now, $\tan \left(\frac{P}{2}+\frac{Q}{2}\right)=1$
$\Rightarrow \frac{\tan \frac{P}{2}+\tan \frac{Q}{2}}{1-\tan \frac{P}{2}\tan \frac{Q}{2}}=1$
$\Rightarrow \frac{-\frac{b}{a}}{1-\frac{c}{a}}=1\Rightarrow -\frac{b}{a}=1-\frac{c}{a}$
$\Rightarrow -b=a-c\Rightarrow c=a+b$
Alternate Solution
$\therefore \angle$$R=\frac{\pi }{2}$
$\Rightarrow \angle P+\angle Q$$=\frac{\pi }{2}$
$\Rightarrow \frac{\angle P}{2}=\frac{\pi }{4}-\frac{\angle Q}{2}$
$\therefore tan\left(\frac{P}{2}\right)=tan\left(\frac{\pi }{4}-\frac{Q}{2}\right)$
$=\frac{\tan \frac{\pi }{4}-\tan \frac{Q}{2}}{1+\tan \frac{\pi }{4}\tan \frac{Q}{2}}$
$\Rightarrow \tan \frac{P}{2}+\tan \frac{P}{2}\tan \frac{Q}{2}=1-\tan \frac{Q}{2}$
$\Rightarrow \tan \frac{P}{2}+\tan \frac{Q}{2}=1-\tan \frac{P}{2}\tan \frac{Q}{2}$
$\Rightarrow -\frac{b}{a}=1-\frac{c}{a}\Rightarrow -b=a-c$
$\Rightarrow c=a+b$