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