In $\triangle \mathrm{ABC}$, with tustal notations, $\mathrm{m} \angle \mathrm{C}=\frac{\pi}{2}$, if $\tan…

In $\triangle \mathrm{ABC}$, with tustal notations, $\mathrm{m} \angle \mathrm{C}=\frac{\pi}{2}$, if $\tan \left(\frac{\mathrm{A}}{2}\right)$ and $\tan \left(\frac{\mathrm{B}}{2}\right)$ are the roots of the equation. $\mathrm{a}_1 x^2+\mathrm{b}_1 x+\mathrm{c}_1=0\left(\mathrm{a}_1 \neq 0\right)$, then
  1. $a_1+b_1=c_1$
  2. $b_1+c_1=a_1$
  3. $\mathrm{a}_1+\mathrm{c}_1-\mathrm{b}_1$
  4. $b_1=c_1$

Solution

$\begin{array}{ll} & \text { In } \triangle \mathrm{ABC}, \\ & \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ} \\ \therefore \quad & \angle \mathrm{A}+\frac{\pi}{2}+\angle \mathrm{B}=180^{\circ} \\ \therefore \quad & \angle \mathrm{A}+\angle \mathrm{B}=\frac{\pi}{2} \\ \therefore \quad & \frac{\angle \mathrm{A}}{2}+\frac{\angle \mathrm{B}}{2}=\frac{\pi}{4} \end{array}$ $\tan \left(\frac{A}{2}\right)$ and $\tan \left(\frac{B}{2}\right)$ are roots of equation $\mathrm{a}_1 x^2+\mathrm{b}_1 x+\mathrm{c}_1=0 ... [Given]$ $\begin{aligned} & \therefore \quad \text { Sum of roots }=\frac{-b_1}{a_1} \\ & \tan \left(\frac{A}{2}\right)+\tan \left(\frac{B}{2}\right)=\frac{-b_1}{a_1}\end{aligned}$ Also, $\tan \left(\frac{\mathrm{A}}{2}\right) \cdot \tan \left(\frac{\mathrm{B}}{2}\right)=\frac{\mathrm{c}_1}{\mathrm{a}_1}$ Using $\tan \left(\frac{\mathrm{A}}{2}+\frac{\mathrm{B}}{2}\right)=\frac{\tan \frac{\mathrm{A}}{2}+\tan \frac{\mathrm{B}}{2}}{1-\tan \frac{\mathrm{A}}{2} \tan \frac{\mathrm{B}}{2}}$, we get $\begin{aligned} & \tan \left(\frac{\pi}{4}\right)=\frac{\frac{-b_1}{a_1}}{1-\frac{c_1}{a_1}} \\ & 1=\frac{-b_1}{a_1-c_1} \\ & a_1-c_1=-b_1 \\ & a_1+b_1=c_1 \end{aligned}$

Asked in: MHT CET 2023 (09 May Shift 2)

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