Let `-pi/6 < theta < -pi/12`. Suppose `alpha_1 and beta_1`, are the roots of the equation `x^2-2xsectheta + 1=0` and `alpha_2 and beta_2` are the roots of the equation `x^2 + 2xtantheta-1=0`. If `alpha_1 > beta_1` and `alpha_2 >beta_2`, then `alpha_1 + beta_2` equals:

Options

A. (a) `2(sec theta-tan theta)`
B. (b) `2 sec theta`
C. (c) `-2 tan theta`
D. (d) `0`

Explanation

To solve the problem step by step, we will analyze the two quadratic equations given and find the roots accordingly.

### Step 1: Analyze the first equation
The first equation is given by:
\[ x^2 - 2x \sec \theta + 1 = 0 \]

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we identify \( a = 1 \), \( b = -2 \sec \theta \), and \( c = 1 \).

### Step 2: Calculate the roots of the first equation
Substituting the values into the quadratic formula:
\[ x = \frac{2 \sec \theta \pm \sqrt{(-2 \sec \theta)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ = \frac{2 \sec \theta \pm \sqrt{4 \sec^2 \theta - 4}}{2} \]
\[ = \sec \theta \pm \sqrt{\sec^2 \theta - 1} \]
Since \( \sqrt{\sec^2 \theta - 1} = \tan \theta \), we can simplify further:
\[ x = \sec \theta \pm \tan \theta \]
Thus, the roots are:
\[ \alpha_1 = \sec \theta + \tan \theta \quad \text{and} \quad \beta_1 = \sec \theta - \tan \theta \]

### Step 3: Analyze the second equation
The second equation is:
\[ x^2 + 2x \tan \theta - 1 = 0 \]

Again, using the quadratic formula with \( a = 1 \), \( b = 2 \tan \theta \), and \( c = -1 \):
\[ x = \frac{-2 \tan \theta \pm \sqrt{(2 \tan \theta)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ = \frac{-2 \tan \theta \pm \sqrt{4 \tan^2 \theta + 4}}{2} \]
\[ = -\tan \theta \pm \sqrt{\tan^2 \theta + 1} \]
Since \( \sqrt{\tan^2 \theta + 1} = \sec \theta \), we can simplify further:
\[ x = -\tan \theta \pm \sec \theta \]
Thus, the roots are:
\[ \alpha_2 = -\tan \theta + \sec \theta \quad \text{and} \quad \beta_2 = -\tan \theta - \sec \theta \]

### Step 4: Calculate \( \alpha_1 + \beta_2 \)
Now, we need to find \( \alpha_1 + \beta_2 \):
\[ \alpha_1 + \beta_2 = (\sec \theta + \tan \theta) + (-\tan \theta - \sec \theta) \]
\[ = \sec \theta + \tan \theta - \tan \theta - \sec \theta \]
\[ = 0 \]

### Final Answer
Thus, \( \alpha_1 + \beta_2 = 0 \).