Question

\( 2\cot^2\theta - \cot\theta - 3 = \) (The question implies factorization of the quadratic in \(\cot\theta\))

• A. \( (2\cot\theta - 3)(\cot\theta + 1) \)
• B. \( (2\cot\theta - 1)(\cot\theta + 3) \)
• C. \( (2\cot\theta + 3)(\cot\theta - 1) \)
• D. \( (2\cot\theta + 1)(\cot\theta - 3) \)

Hint:

To factor a quadratic \(ax^2+bx+c\), find two numbers that multiply to \(ac\) and add to \(b\).
Alternatively, you can expand the given options to see which one matches the original expression.
Let \(y = \cot\theta\) to simplify the appearance of the quadratic before factoring.

Answer 1

The Correct Option is A

Let \(y = \cot\theta\). The expression becomes a quadratic in y: \( 2y^2 - y - 3 \) We need to factorize this quadratic. We are looking for two numbers whose product is \(2 \times (-3) = -6\) and whose sum is \(-1\) (the coefficient of y). The numbers are \(-3\) and \(2\), since \((-3) \times 2 = -6\) and \(-3 + 2 = -1\). Rewrite the middle term using these numbers: \( 2y^2 - 3y + 2y - 3 \) Factor by grouping: \( y(2y - 3) + 1(2y - 3) \) \( (2y - 3)(y + 1) \) Now substitute back \(y = \cot\theta\): \( (2\cot\theta - 3)(\cot\theta + 1) \) This matches option (a). Alternatively, expand the options: % Option (a) \((2\cot\theta - 3)(\cot\theta + 1) = 2\cot^2\theta + 2\cot\theta - 3\cot\theta - 3 = 2\cot^2\theta - \cot\theta - 3\). (Matches) % Option (b) \((2\cot\theta - 1)(\cot\theta + 3) = 2\cot^2\theta + 6\cot\theta - \cot\theta - 3 = 2\cot^2\theta + 5\cot\theta - 3\). (No) % Option (c) \((2\cot\theta + 3)(\cot\theta - 1) = 2\cot^2\theta - 2\cot\theta + 3\cot\theta - 3 = 2\cot^2\theta + \cot\theta - 3\). (No) % Option (d) \((2\cot\theta + 1)(\cot\theta - 3) = 2\cot^2\theta - 6\cot\theta + \cot\theta - 3 = 2\cot^2\theta - 5\cot\theta - 3\). (No) \[ \boxed{(2\cot\theta - 3)(\cot\theta + 1)} \]