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In a triangle $P Q R, \angle R=\frac{\pi}{2} .$ If $\tan \left(\frac{P}{2}\right)$ and $\tan \left(\frac{Q}{2}\right)$ are the roots of $a x^{2}+\bar{b} x+c=0$ $a \neq 0$ then

1) $a=b+c$
2) $c=a+b$
3) $b=c$
4) $b=a+c$
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(2)

Explanation

$\tan \left(\frac{P}{2}\right), \tan \left(\frac{Q}{2}\right)$ are the roots of

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

and $\tan \left(\frac{P}{2}\right) \times \tan \left(\frac{Q}{2}\right)=\frac{c}{a}$

$\Rightarrow \frac{\tan \left(\frac{P}{2}\right)+\tan \left(\frac{Q}{2}\right)}{1-\tan \left(\frac{P}{2}\right)+\tan \left(\frac{Q}{2}\right)}=\tan \left(\frac{P}{2}+\frac{Q}{2}\right)=\tan 45^{0}=1$

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

$c=a+b$

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