For any real numbers `theta` and `phi` we define `theta R phi` if and only if `sec^(2) theta - tan^(2) phi =1` the relation R is

Options

A. reflexive but not transitive
B. symmetric but not reflexive
C. both reflexive and symmetric but not transitive
D. an equivalence relation

Answer

To solve the problem, we need to analyze the relation defined by \( \theta R \phi \) if and only if \( \sec^2 \theta - \tan^2 \phi = 1 \). We will determine the nature of this relation.

### Step 1: Understanding the relation
The relation is defined as:
\[ \theta R \phi \iff \sec^2 \theta - \tan^2 \phi = 1 \]

### Step 2: Using the trigonometric identity
We know from trigonometric identities that:
\[ \sec^2 \theta - \tan^2 \theta = 1 \]
This means that if we let \( \phi = \theta \), then:
\[ \sec^2 \theta - \tan^2 \theta = 1 \]
Thus, \( \theta R \theta \) holds true.

### Step 3: Checking reflexivity
For a relation to be reflexive, it must hold that \( \theta R \theta \) for all \( \theta \). Since we have shown that:
\[ \sec^2 \theta - \tan^2 \theta = 1 \]
is true, the relation is reflexive.

### Step 4: Checking symmetry
For symmetry, we need to check if \( \theta R \phi \) implies \( \phi R \theta \). Given that:
\[ \sec^2 \theta - \tan^2 \phi = 1 \]
and if we assume \( \theta = \phi \), then:
\[ \sec^2 \phi - \tan^2 \theta = 1 \]
Thus, the relation is symmetric.

### Step 5: Checking transitivity
For transitivity, we need to check if \( \theta R \phi \) and \( \phi R \psi \) imply \( \theta R \psi \). If we have:
\[ \sec^2 \theta - \tan^2 \phi = 1 \quad \text{and} \quad \sec^2 \phi - \tan^2 \psi = 1 \]
This means that \( \theta \) must equal \( \phi \) and \( \phi \) must equal \( \psi \). Therefore, \( \theta \) must equal \( \psi \), which implies:
\[ \sec^2 \theta - \tan^2 \psi = 1 \]
Thus, the relation is transitive.

### Conclusion
Since the relation \( R \) is reflexive, symmetric, and transitive, it is an equivalence relation.

The final answer is that the relation \( R \) is an equivalence relation.

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