If z = `sec^(-1) (x + 1/x) + sec^(-1) (y + 1/y)`, where xy< 0, then the possible value of z is (are)

Options

A. `(8pi)/(10)`
B. `(7pi)/(10)`
C. `(9pi)/(10)`
D. `(21pi)/(20)`

Explanation

To solve the problem, we need to analyze the expression given for \( z \):

\[ z = \sec^{-1} \left( x + \frac{1}{x} \right) + \sec^{-1} \left( y + \frac{1}{y} \right) \]

where \( xy < 0 \).

### Step 1: Analyze the expressions \( x + \frac{1}{x} \) and \( y + \frac{1}{y} \)

The expression \( x + \frac{1}{x} \) has certain properties based on the value of \( x \):
- If \( x > 0 \), then \( x + \frac{1}{x} \geq 2 \).
- If \( x < 0 \), then \( x + \frac{1}{x} \leq -2 \).

Similarly, for \( y \):
- If \( y > 0 \), then \( y + \frac{1}{y} \geq 2 \).
- If \( y < 0 \), then \( y + \frac{1}{y} \leq -2 \).

Given that \( xy < 0 \), one of \( x \) or \( y \) must be positive and the other must be negative.

### Step 2: Choose combinations based on the sign of \( x \) and \( y \)

Since \( xy < 0 \), we can choose:
- \( x + \frac{1}{x} \geq 2 \) (for \( x > 0 \))
- \( y + \frac{1}{y} \leq -2 \) (for \( y < 0 \))

### Step 3: Determine the ranges for \( z \)

Now we can find the ranges for the inverse secant functions:

1. For \( \sec^{-1} \left( x + \frac{1}{x} \right) \):
- Since \( x + \frac{1}{x} \geq 2 \), the range of \( \sec^{-1} \) will be:
\[ \sec^{-1} \left( x + \frac{1}{x} \right) \in \left[ \frac{\pi}{3}, \frac{\pi}{2} \right] \]

2. For \( \sec^{-1} \left( y + \frac{1}{y} \right) \):
- Since \( y + \frac{1}{y} \leq -2 \), the range of \( \sec^{-1} \) will be:
\[ \sec^{-1} \left( y + \frac{1}{y} \right) \in \left[ \frac{\pi}{2}, \frac{2\pi}{3} \right] \]

### Step 4: Combine the ranges to find \( z \)

Now we can add the ranges of the two inverse secant functions to find the range for \( z \):

- The minimum value of \( z \):
\[ \frac{\pi}{3} + \frac{\pi}{2} = \frac{2\pi}{6} + \frac{3\pi}{6} = \frac{5\pi}{6} \]

- The maximum value of \( z \):
\[ \frac{\pi}{2} + \frac{2\pi}{3} = \frac{3\pi}{6} + \frac{4\pi}{6} = \frac{7\pi}{6} \]

Thus, the range of \( z \) is:
\[ z \in \left[ \frac{5\pi}{6}, \frac{7\pi}{6} \right] \]

### Step 5: Identify possible values of \( z \)

Now we need to check which values fall within this range. The only value that lies in this interval is \( \frac{9\pi}{10} \).

### Final Answer

The possible value of \( z \) is:
\[ \frac{9\pi}{10} \]