Question

Factorize \[ 2\cot^2\theta-\cot\theta-3 \]

• 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:

For quadratic expressions of the form \[ ax^2+bx+c, \] find two numbers whose product is \(ac\) and whose sum is \(b\). Then use factorization by grouping.

Answer 1

The Correct Option is A

Step 1: Substitute a variable.
Let \[ x=\cot\theta \] Then the expression becomes \[ 2x^2-x-3 \]

Step 2: Split the middle term.
We need two numbers whose product is \[ 2\times(-3)=-6 \] and whose sum is \[ -1 \] The required numbers are \[ -3 \quad \text{and} \quad 2 \] Therefore, \[ 2x^2-x-3 = 2x^2+2x-3x-3 \]

Step 3: Factor by grouping.
\[ = 2x(x+1)-3(x+1) \] \[ = (x+1)(2x-3) \] Substituting back \(x=\cot\theta\), \[ = (\cot\theta+1)(2\cot\theta-3) \]

Step 4: Verification.
Expanding, \[ (2\cot\theta-3)(\cot\theta+1) \] \[ = 2\cot^2\theta+2\cot\theta -3\cot\theta-3 \] \[ = 2\cot^2\theta-\cot\theta-3 \] which matches the given expression.

Step 5: Final conclusion.
Hence, \[ \boxed{(2\cot\theta-3)(\cot\theta+1)} \] Therefore, the correct option is \[ \boxed{(1)\ (2\cot\theta-3)(\cot\theta+1)} \]