Examine whether the function are even or odd or none.
`f(x) = (sec x + x^(2) -9)/(x sin x )`

Answer

To determine whether the function \( f(x) = \frac{\sec x + x^2 - 9}{x \sin x} \) is even, odd, or neither, we will follow these steps:

### Step 1: Define the function
We have the function:
\[ f(x) = \frac{\sec x + x^2 - 9}{x \sin x} \]

### Step 2: Find \( f(-x) \)
To check if the function is even or odd, we need to compute \( f(-x) \):
\[ f(-x) = \frac{\sec(-x) + (-x)^2 - 9}{(-x) \sin(-x)} \]

### Step 3: Simplify \( f(-x) \)
Using the properties of trigonometric functions:
- \( \sec(-x) = \sec x \) (since secant is an even function)
- \( (-x)^2 = x^2 \)
- \( \sin(-x) = -\sin x \)

Substituting these into \( f(-x) \):
\[ f(-x) = \frac{\sec x + x^2 - 9}{-x \sin x} \]
This can be rewritten as:
\[ f(-x) = -\frac{\sec x + x^2 - 9}{x \sin x} = -f(x) \]

### Step 4: Conclusion
Since \( f(-x) = -f(x) \), the function \( f(x) \) is an **odd function**.

### Summary
We have determined that the function \( f(x) = \frac{\sec x + x^2 - 9}{x \sin x} \) is an odd function.

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