Question

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

Hint:

Use trigonometric identities like \( \cot(x + y) \) and \( \cot(x) \cot(y) \) to derive relations for \( \cot x + \cot y \) and form the quadratic equation.

Answer 1

The Correct Option is B

We are given that: \[ \cot x \cot y = a \quad \text{and} \quad x + y = \frac{\pi}{6} \] From the identity for \( \cot(x + y) \): \[ \cot(x + y) = \frac{\cot x \cot y - 1}{\cot x + \cot y} \] Substituting \( x + y = \frac{\pi}{6} \), we get: \[ \cot \left( \frac{\pi}{6} \right) = \frac{\cot x \cot y - 1}{\cot x + \cot y} \] Since \( \cot \left( \frac{\pi}{6} \right) = \sqrt{3} \), this becomes: \[ \sqrt{3} = \frac{a - 1}{\cot x + \cot y} \] Let \( t = \cot x + \cot y \). Therefore, we have the equation: \[ \sqrt{3}t = a - 1 \] Solving for \( t \), we get: \[ t = \frac{a - 1}{\sqrt{3}} \] Step 1: To form the quadratic equation with roots \( \cot x \) and \( \cot y \), we use the standard form: \[ t^2 - (\cot x + \cot y)t + \cot x \cot y = 0 \] Substituting \( \cot x \cot y = a \) and \( \cot x + \cot y = t \), we get: \[ t^2 - t + a = 0 \] This simplifies to: \[ \sqrt{3}t^2 + (1 - a)t + a\sqrt{3} = 0 \] % Final Answer \[ \boxed{\sqrt{3}t^2 + (1 - a)t + a\sqrt{3} = 0} \]

Answer 2

Given:
- \( \cot x \cot y = a \)
- \( x + y = \frac{\pi}{6} \)

We are to find the quadratic equation whose roots are \( \cot x \) and \( \cot y \).

Step 1: Use identity for sum of cotangents
From the identity for cotangent of sum of angles:
\[ \cot(x + y) = \frac{\cot x \cot y - 1}{\cot x + \cot y} \]
Let \( \cot x = t_1 \), \( \cot y = t_2 \)
Then: - \( t_1 + t_2 = S \) (sum of roots)
- \( t_1 t_2 = a \) (product of roots)

Given:
\[ \cot(x + y) = \cot\left(\frac{\pi}{6}\right) = \sqrt{3} \]
Apply identity:
\[ \cot(x + y) = \frac{t_1 t_2 - 1}{t_1 + t_2} = \frac{a - 1}{S} \]
So, \[ \frac{a - 1}{S} = \sqrt{3} \Rightarrow S = \frac{a - 1}{\sqrt{3}} \]

Step 2: Form the quadratic equation
Quadratic with roots \( t_1 \) and \( t_2 \) is:
\[ t^2 - (t_1 + t_2)t + t_1 t_2 = 0 \Rightarrow t^2 - S t + a = 0 \]
Substitute \( S = \frac{a - 1}{\sqrt{3}} \):
\[ t^2 - \left( \frac{a - 1}{\sqrt{3}} \right)t + a = 0 \]
Multiply entire equation by \( \sqrt{3} \) to eliminate denominator:
\[ \sqrt{3}t^2 + (1 - a)t + a\sqrt{3} = 0 \]

Final Answer:
\[ \boxed{\sqrt{3}t^2 + (1 - a)t + a\sqrt{3} = 0} \]