Finding cot(α + β) from Quadratic Roots

The problem provides a quadratic equation \(\text{x}^2 - 3\text{x} + 2 = 0\) and states that its roots are \(\text{cot } \alpha\) and \(\text{cot } \beta\). We need to find the value of \(\text{cot}(\alpha + \beta)\).

Analyzing the Quadratic Equation

The given quadratic equation is \(\text{x}^2 - 3\text{x} + 2 = 0\). We can find the roots of this equation by factoring or using the quadratic formula. Factoring the equation:

\((\text{x} - 1)(\text{x} - 2) = 0\)

The roots are \(\text{x} = 1\) and \(\text{x} = 2\).

Since the roots are \(\text{cot } \alpha\) and \(\text{cot } \beta\), we can say that \(\{\text{cot } \alpha, \text{cot } \beta\} = \{1, 2\}\). It doesn't matter which root is \(\text{cot } \alpha\) and which is \(\text{cot } \beta\), as the expression we need to evaluate, \(\text{cot}(\alpha + \beta)\), is symmetric with respect to \(\alpha\) and \(\beta\).

Using Vieta's Formulas (Sum and Product of Roots)

Alternatively, we can use Vieta's formulas, which relate the roots of a quadratic equation to its coefficients. For a quadratic equation \(\text{ax}^2 + \text{bx} + \text{c} = 0\), the sum of the roots is \(-\text{b}/\text{a}\) and the product of the roots is \(\text{c}/\text{a}\).

In our equation, \(\text{x}^2 - 3\text{x} + 2 = 0\), we have \(\text{a} = 1\), \(\text{b} = -3\), and \(\text{c} = 2\).

• Sum of roots: \(\text{cot } \alpha + \text{cot } \beta = -(-3)/1 = 3\).
• Product of roots: \((\text{cot } \alpha)(\text{cot } \beta) = 2/1 = 2\).

These values (sum = 3, product = 2) are consistent with the roots we found by factoring (1 + 2 = 3, 1 × 2 = 2).

Applying the Cotangent Sum Formula

We need to find \(\text{cot}(\alpha + \beta)\). The formula for the cotangent of the sum of two angles is:

\(\text{cot}(\alpha + \beta) = \frac{\text{cot } \alpha \text{ cot } \beta - 1}{\text{cot } \alpha + \text{cot } \beta}\)

Substituting Values and Calculating

Now, we substitute the sum and product of the roots we found into the cotangent sum formula:

\(\text{cot}(\alpha + \beta) = \frac{(\text{cot } \alpha)(\text{cot } \beta) - 1}{\text{cot } \alpha + \text{cot } \beta}\)

Substitute \((\text{cot } \alpha)(\text{cot } \beta) = 2\) and \(\text{cot } \alpha + \text{cot } \beta = 3\):

\(\text{cot}(\alpha + \beta) = \frac{2 - 1}{3}\)

\(\text{cot}(\alpha + \beta) = \frac{1}{3}\)

Therefore, the value of \(\text{cot}(\alpha + \beta)\) is \(\frac{1}{3}\).

Revision Table: Key Concepts

Additional Information: Trigonometric Identities

Understanding trigonometric identities is crucial for solving problems involving trigonometric functions. The cotangent sum formula used here is derived from the tangent sum formula. Recall the relationship \(\text{cot } \theta = 1 / \text{tan } \theta\).

The tangent sum formula is:

\(\text{tan}(\alpha + \beta) = \frac{\text{tan } \alpha + \text{tan } \beta}{1 - \text{tan } \alpha \text{ tan } \beta}\)

To get the cotangent formula, we take the reciprocal of the tangent formula:

\(\text{cot}(\alpha + \beta) = \frac{1}{\text{tan}(\alpha + \beta)} = \frac{1 - \text{tan } \alpha \text{ tan } \beta}{\text{tan } \alpha + \text{tan } \beta}\)

Now, substitute \(\text{tan } \alpha = 1 / \text{cot } \alpha\) and \(\text{tan } \beta = 1 / \text{cot } \beta\):

\(\text{cot}(\alpha + \beta) = \frac{1 - (1 / \text{cot } \alpha)(1 / \text{cot } \beta)}{(1 / \text{cot } \alpha) + (1 / \text{cot } \beta)}\)

Multiply the numerator and denominator by \((\text{cot } \alpha)(\text{cot } \beta)\):

\(\text{cot}(\alpha + \beta) = \frac{(\text{cot } \alpha)(\text{cot } \beta) \left(1 - \frac{1}{\text{cot } \alpha \text{ cot } \beta}\right)}{(\text{cot } \alpha)(\text{cot } \beta) \left(\frac{1}{\text{cot } \alpha} + \frac{1}{\text{cot } \beta}\right)}\)

\(\text{cot}(\alpha + \beta) = \frac{\text{cot } \alpha \text{ cot } \beta - 1}{\text{cot } \beta + \text{cot } \alpha}\)

This confirms the formula used in the solution. This problem elegantly combines concepts from quadratic equations (roots and coefficients) and trigonometry (sum identities).