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In $\Delta \mathrm{PQR}, \angle \mathrm{R}=\frac{\pi}{2}$. If than $\left(\frac{1}{2}\right)$ and
$\left(\frac{Q}{2}\right)$ are the roots of the equation $a x^{2}+b x+c=0$, then which one of the following is correct?
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$\left(\frac{Q}{2}\right)$ are the roots of the equation $a x^{2}+b x+c=0$, then which one of the following is correct?
Solution:
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Verified Answer
The correct answer is:
$\mathrm{c}=\mathrm{a}+\mathrm{b}$
$\angle \mathrm{R}=\frac{\pi}{2}$
$\therefore \angle \mathrm{P}+\angle \mathrm{Q}=\frac{\pi}{2}$
$\Rightarrow \frac{\angle \mathrm{P}+\angle \mathrm{Q}}{2}=\frac{\pi}{4}$...(1)
$\tan \left(\frac{P}{2}+\frac{Q}{2}\right)=\frac{\tan \frac{P}{2}+\tan \frac{Q}{2}}{1-\tan \frac{P}{2} \tan \frac{Q}{2}}$...(2)
Given, $\tan \frac{P}{2}$ and $\tan \frac{Q}{2}$ are roots of $a x^{2}+b x+c=0$. $\tan \frac{P}{2}+\tan \frac{Q}{2}=\frac{-b}{a} ; \tan \frac{P}{2} \cdot \tan \frac{Q}{2}=\frac{c}{a}$
$(2) \Rightarrow \tan \left(\frac{\mathrm{P}}{2}+\frac{\mathrm{Q}}{2}\right)=\frac{\frac{-\mathrm{b}}{\mathrm{a}}}{1-\frac{\mathrm{c}}{\mathrm{a}}}$
$\Rightarrow \tan \frac{\pi}{4}=\frac{\frac{-b}{a}}{1-\frac{c}{a}} \quad($ from $(1))$
$\Rightarrow 1=\frac{-b}{a-c} \Rightarrow-b=a-c \Rightarrow a+b=c$
$11-x+z^{2}=5$
$\therefore \angle \mathrm{P}+\angle \mathrm{Q}=\frac{\pi}{2}$
$\Rightarrow \frac{\angle \mathrm{P}+\angle \mathrm{Q}}{2}=\frac{\pi}{4}$...(1)
$\tan \left(\frac{P}{2}+\frac{Q}{2}\right)=\frac{\tan \frac{P}{2}+\tan \frac{Q}{2}}{1-\tan \frac{P}{2} \tan \frac{Q}{2}}$...(2)
Given, $\tan \frac{P}{2}$ and $\tan \frac{Q}{2}$ are roots of $a x^{2}+b x+c=0$. $\tan \frac{P}{2}+\tan \frac{Q}{2}=\frac{-b}{a} ; \tan \frac{P}{2} \cdot \tan \frac{Q}{2}=\frac{c}{a}$
$(2) \Rightarrow \tan \left(\frac{\mathrm{P}}{2}+\frac{\mathrm{Q}}{2}\right)=\frac{\frac{-\mathrm{b}}{\mathrm{a}}}{1-\frac{\mathrm{c}}{\mathrm{a}}}$
$\Rightarrow \tan \frac{\pi}{4}=\frac{\frac{-b}{a}}{1-\frac{c}{a}} \quad($ from $(1))$
$\Rightarrow 1=\frac{-b}{a-c} \Rightarrow-b=a-c \Rightarrow a+b=c$
$11-x+z^{2}=5$
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