In a $\triangle \mathrm{PQR}, \mathrm{m} \angle \mathrm{R}=\frac{\pi}{2}$. If $\tan…

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

Solution

$\begin{aligned} & \text { In } \triangle \mathrm{PQR} \\ & \angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}=180^{\circ} \\ \therefore \quad & \angle \mathrm{P}+\angle \mathrm{Q}+\frac{\pi}{2}=180^{\circ} \\ \therefore \quad & \angle \mathrm{P}+\angle \mathrm{Q}=\frac{\pi}{2} \\ \therefore \quad & \frac{\angle \mathrm{P}}{2}+\frac{\angle \mathrm{Q}}{2}=\frac{\pi}{4} \end{aligned}$ $\tan \left(\frac{\mathrm{P}}{2}\right)$ and $\tan \left(\frac{\mathrm{Q}}{2}\right)$ are roots of the equation $a x^2+b x+c=0$...[Given] $\therefore \quad$ Sum of roots $=\frac{-b}{a}$ $\tan \left(\frac{\mathrm{P}}{2}\right)+\tan \left(\frac{\mathrm{Q}}{2}\right)=\frac{-\mathrm{b}}{\mathrm{a}}$ Also, $\tan \left(\frac{\mathrm{P}}{2}\right) \cdot \tan \left(\frac{\mathrm{Q}}{2}\right)=\frac{\mathrm{c}}{\mathrm{a}}$ Using, $\tan \left(\frac{P}{2}+\frac{Q}{2}\right)=\frac{\tan \frac{P}{2}+\tan \frac{Q}{2}}{1-\tan \frac{P}{2} \tan \frac{Q}{2}}$, we get $\begin{aligned} & \Rightarrow \tan \left(\frac{\pi}{4}\right)=\frac{\frac{-b}{a}}{1-\frac{c}{a}} \\ & \Rightarrow 1=\frac{-b}{a-c} \\ & \Rightarrow \mathrm{a}-\mathrm{c}=-\mathrm{b} \\ & \Rightarrow \mathrm{a}+\mathrm{b}=\mathrm{c} \end{aligned}$

Asked in: MHT CET 2024 (11 May Shift 2)

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