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 $ax^{2} + bx + c = 0, a\ne0$, 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 $ax^{2} + bx + c = 0.$
$\therefore \tan \frac{P}{2}+\tan \frac{Q}{2}=-\frac{b}{a}$
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}$
$\Rightarrow \frac{\angle P}{2}=\frac{\pi}{4}-\frac{\angle Q}{2}$
$\therefore tan\left(\frac{P}{2}\right)=tan\left(\frac{\pi}{4}-\frac{Q}{2}\right)$
$=\frac{\tan \frac{\pi }{4}-\tan\frac{Q }{2}}{1+\tan \frac{\pi }{4} \tan\frac{Q}{2}}$
$\Rightarrow \tan \frac{P }{2}+\tan \frac{P}{2}\tan \frac{Q}{2}=1-\tan \frac{Q}{2}$
$\Rightarrow \tan \frac{P}{2}+\tan \frac{Q}{2}=1-\tan \frac{P}{2} \tan \frac{Q}{2}$
$\Rightarrow -\frac{b}{a}=1-\frac{c}{a} \Rightarrow -b=a-c$
$\Rightarrow c=a+b$