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If cot α and cot β are the roots of the equation x 2+ bx + c = 0 with b ≠ 0, then the value of cot (α + β) is

This question was previously asked in
NDA I 2017 GAT Previous Year Paper (23-Apr-2017)
The correct answer is \(\frac{{1 - {\rm{c}}}}{{\rm{b}}}\)

We are given a quadratic equation \({x^2 + bx + c = 0}\) where \({b \neq 0}\). The roots of this equation are given as \({cot \alpha}\) and \({cot \beta}\).

For a quadratic equation of the form \({Ax^2 + Bx + C = 0}\), the sum of the roots is given by \({-B/A}\) and the product of the roots is given by \({C/A}\).

In our case, the equation is \({x^2 + bx + c = 0}\). Comparing this to \({Ax^2 + Bx + C = 0}\), we have \({A=1}\), \({B=b}\), and \({C=c}\). The roots are \({x_1 = \cot \alpha}\) and \({x_2 = \cot \beta}\).

Relating Roots and Coefficients

Using the properties of the roots of a quadratic equation:

  • Sum of roots: \({\cot \alpha + \cot \beta = \frac{-b}{1} = -b}\)
  • Product of roots: \({(\cot \alpha)(\cot \beta) = \frac{c}{1} = c}\)

Using the Cotangent Addition Formula

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

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

Now, we can substitute the values we found for the sum and product of the roots into this formula.

We have \({\cot \alpha + \cot \beta = -b}\) and \({(\cot \alpha)(\cot \beta) = c}\).

Substituting these values into the formula for \({\cot(\alpha + \beta)}\):

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

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

To simplify and remove the negative sign in the denominator, we can multiply both the numerator and the denominator by \({-1}\):

\({\cot(\alpha + \beta) = \frac{(c - 1) \times (-1)}{(-b) \times (-1)}}\)

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

This can also be written as:

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

Calculation Summary

Property Value from Equation
Sum of roots (\({cot \alpha + cot \beta}\)) \({-b}\)
Product of roots (\({(cot \alpha)(cot \beta)}\)) \({c}\)
Identity for \({cot(\alpha + \beta)}\) \({\frac{cot \alpha cot \beta - 1}{cot \alpha + cot \beta}}\)
Calculated \({cot(\alpha + \beta)}\) \({\frac{c - 1}{-b} = \frac{1 - c}{b}}\)

Therefore, the value of \({\cot(\alpha + \beta)}\) is \({\frac{1 - c}{b}}\).

Revision Table: Quadratic Roots and Trigonometry

Concept Description Formula/Relation
Roots of Quadratic Eq. For \({Ax^2 + Bx + C = 0}\), roots \({x_1, x_2}\) Sum of roots: \({x_1 + x_2 = -B/A}\)
Product of roots: \({x_1 x_2 = C/A}\)
Cotangent Addition Formula for \({cot}\) of sum of angles \({\cot(A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B}}\)

Additional Information: Connecting Algebra and Trigonometry

This problem is a good example of how concepts from algebra (properties of quadratic equations) and trigonometry (identities) can be combined. When roots of a quadratic equation are given in terms of trigonometric functions, you should always consider using the sum and product of roots relations and relevant trigonometric identities. The condition \({b \neq 0}\) is important because if \({b=0}\), the sum of roots \({cot \alpha + cot \beta = 0}\), which might lead to an undefined value in the denominator of the \({cot(\alpha + \beta)}\) formula if \({cot \alpha + cot \beta = 0}\) is the denominator, however, here the denominator is \(-b\), so \({b \neq 0}\) ensures the denominator \(-b\) is not zero.

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