The least value of |a| for which `tan theta` and `cot theta` are the roots of the equation `x^(2)+ax+b=0`

Options

A. 2
B. 1
C. `1//2`
D. 0

Answer

To find the least value of |a| for which tan(θ) and cot(θ) are the roots of the equation \(x^2 + ax + b = 0\), we can follow these steps:

### Step 1: Identify the roots
The roots of the quadratic equation are given as:
- \(\alpha = \tan(\theta)\)
- \(\beta = \cot(\theta)\)

### Step 2: Use the relationship between roots and coefficients
From Vieta's formulas, we know:
- The sum of the roots \(\alpha + \beta = -a\)
- The product of the roots \(\alpha \beta = b\)

### Step 3: Calculate the sum of the roots
We can express the sum of the roots:
\[ \tan(\theta) + \cot(\theta) = \tan(\theta) + \frac{1}{\tan(\theta)} = \frac{\tan^2(\theta) + 1}{\tan(\theta)} \]
Using the identity \(1 + \tan^2(\theta) = \sec^2(\theta)\), we have:
\[ \tan(\theta) + \cot(\theta) = \frac{\sec^2(\theta)}{\tan(\theta)} \]

### Step 4: Simplify the expression for a
Thus, we can express \(a\) as:
\[ -a = \tan(\theta) + \cot(\theta) \implies a = -\left(\tan(\theta) + \cot(\theta)\right) \]

### Step 5: Substitute and simplify
Substituting the expression for \(\tan(\theta) + \cot(\theta)\):
\[ a = -\left(\tan(\theta) + \frac{1}{\tan(\theta)}\right) \]
Let \(x = \tan(\theta)\), then:
\[ a = -\left(x + \frac{1}{x}\right) \]

### Step 6: Find the minimum value of |a|
We need to minimize \(|a|\):
\[ |a| = \left| -\left(x + \frac{1}{x}\right) \right| = \left| x + \frac{1}{x} \right| \]
The expression \(x + \frac{1}{x}\) is minimized when \(x = 1\) (using AM-GM inequality), giving:
\[ x + \frac{1}{x} \geq 2 \]
Thus, the minimum value of \(|a|\) occurs at:
\[ |a| = 2 \]

### Conclusion
The least value of \(|a|\) for which \(\tan(\theta)\) and \(\cot(\theta)\) are the roots of the equation \(x^2 + ax + b = 0\) is:
\[ \boxed{2} \]