If $\cot x \cot y=a$ and $x+y=\frac{\pi}{6}$, then the quadratic equation satisfying $\cot \mathrm{x}$ and…

If $\cot x \cot y=a$ and $x+y=\frac{\pi}{6}$, then the quadratic equation satisfying $\cot \mathrm{x}$ and $\cot \mathrm{y}$ is
  1. $t^2+(1-a) \sqrt{3} t+a=0$
  2. $\sqrt{3} t^2+(1-a) t+a \sqrt{3}=0$
  3. $\sqrt{3} t^2+(a-1) t+a \sqrt{3}=0$
  4. $\mathrm{t}^2+(\mathrm{a}-1) \sqrt{3} \mathrm{t}+\mathrm{a}=0$

Solution

$\begin{aligned} & \text {Given } x+y=\frac{\pi}{6} \\ & \Rightarrow \cot (x+y)=\cot \left(\frac{\pi}{6}\right) \\ & \Rightarrow \frac{\cot x \cot y-1}{\cot x+\cot y}=\sqrt{3} \\ & \Rightarrow \cot x+\cot y=\frac{1}{\sqrt{3}}(\cot x \cdot \cot y-1)=\frac{1}{\sqrt{3}}(a-1) \end{aligned}$ So, quadratic equation whose roots are $\cot x, \cot y$ is $\begin{aligned} & t^2-(\cot x+\cot y) t+\cot x \cdot \cot y=0 \\ & \Rightarrow t^2-\frac{1}{\sqrt{3}}(a-1) t+a=0 \\ & \Rightarrow \sqrt{3} t^2+(1-a) t+\sqrt{3} a=0 \end{aligned}$

Asked in: AP EAMCET 2023 (15 May Shift 2)

Practice more Trigonometric Ratios & Identities questions on Aicharya