Let $\tan 30^{\circ}$ and $\tan 15^{\circ}$ be the roots of the quadratic equation $x^2+a x+b=0$, then…

Let $\tan 30^{\circ}$ and $\tan 15^{\circ}$ be the roots of the quadratic equation $x^2+a x+b=0$, then $1+a-b=$
  1. $0$
  2. $1$
  3. $a b$
  4. $a^2 b^2$

Solution

$\begin{aligned} & \tan 15^{\circ}=\tan \left(45^{\circ}-30^{\circ}\right) \\ & =\frac{1-1 / \sqrt{3}}{1+1 / \sqrt{3}}=\frac{\sqrt{3}-1}{\sqrt{3}+1} .\end{aligned}$ Given, $\tan 30^{\circ}$ and $\tan 15^{\circ}$ are the roots of the equation $x^2+a x+b=0$ $\because$ Sum of roots $=\tan 30^{\circ}+\tan 15^{\circ}=-a$ $\begin{aligned} & \Rightarrow \frac{1}{\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}+1}=-a \\ & \Rightarrow \frac{\sqrt{3}+1+3-\sqrt{3}}{\sqrt{3}(\sqrt{3}+1)}=-a \Rightarrow a=\frac{-4}{\sqrt{3}(\sqrt{3}+1)}\end{aligned}$ Now, product of roots $=\tan 30^{\circ} \cdot \tan 15^{\circ}=b$ $\Rightarrow \quad b=\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}-1}{\sqrt{3}+1}$ Now, we have to find $1+a-b$. Then, $1+a-b=1-\frac{4}{\sqrt{3}(\sqrt{3}+1)}-\frac{\sqrt{3}-1}{\sqrt{3}(\sqrt{3}+1)}$ $=\frac{\sqrt{3}(\sqrt{3}+1)-4-\sqrt{3}+1}{\sqrt{3}(\sqrt{3}+1)}=\frac{3+\sqrt{3}-3-\sqrt{3}}{\sqrt{3}(\sqrt{3}+1)}=0$

Asked in: AP EAMCET 2022 (08 Jul Shift 2)

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