AP EAMCET · Maths · Trigonometric Ratios & Identities
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
- A \(t^2+(1-a) \sqrt{3} t+a=0\)
- B \(\sqrt{3} t^2+(1-a) t+a \sqrt{3}=0\)
- C \(\sqrt{3} t^2+(a-1) t+a \sqrt{3}=0\)
- D \(\mathrm{t}^2+(\mathrm{a}-1) \sqrt{3} \mathrm{t}+\mathrm{a}=0\)
Answer & Solution
Correct Answer
(B) \(\sqrt{3} t^2+(1-a) t+a \sqrt{3}=0\)
Step-by-step Solution
Detailed explanation
\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)…
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