Let `f(x) = sin^2(x//2) + cos^2(x//2) and g(x) = sec^2x- tan^2x`. The two function are equal over the set :

Options

A. `phi`
B. R
C. `R - {x : x = (2n + 1)(pi)/(2), n in Z}`
D. `R^(+)`

Answer

To solve the problem, we need to analyze the two functions \( f(x) \) and \( g(x) \) given in the question.

1. **Define the functions**:
\[ f(x) = \sin^2\left(\frac{x}{2}\right) + \cos^2\left(\frac{x}{2}\right) \]
\[ g(x) = \sec^2(x) - \tan^2(x) \]

2. **Simplify \( f(x) \)**:
We know from the Pythagorean identity that:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
Therefore, we can simplify \( f(x) \):
\[ f(x) = 1 \]

3. **Simplify \( g(x) \)**:
We can use the identity \( 1 + \tan^2(x) = \sec^2(x) \) to simplify \( g(x) \):
\[ g(x) = \sec^2(x) - \tan^2(x) = 1 \]

4. **Determine the domain of \( g(x) \)**:
The function \( g(x) \) is not defined where \( \sec(x) \) and \( \tan(x) \) are undefined. This occurs at:
\[ x = \frac{\pi}{2} + n\pi \quad \text{for } n \in \mathbb{Z} \]
This means \( g(x) \) is undefined at odd multiples of \( \frac{\pi}{2} \).

5. **Conclusion**:
The two functions \( f(x) \) and \( g(x) \) are equal when:
\[ f(x) = g(x) = 1 \quad \text{for } x \in \mathbb{R} \setminus \left\{ \frac{\pi}{2} + n\pi \right\} \]
In set notation, this can be expressed as:
\[ x \in \mathbb{R} \setminus \left\{ x = \frac{\pi}{2} + n\pi \, | \, n \in \mathbb{Z} \right\} \]

### Final Answer:
The two functions are equal over the set:
\[ \mathbb{R} \setminus \left\{ x = \frac{\pi}{2} + n\pi \, | \, n \in \mathbb{Z} \right\} \]