`sec theta=(a^(2)+b^(2))/(a^(2)-b^(2)), where a, binR,` gives real balues of `theta` if and only if

Options

A. `a=bne0`
B. `|a|ne|b|ne0`
C. `a+b=0,ane0`
D. none of these

Answer

To solve the problem, we need to analyze the given equation \( \sec \theta = \frac{a^2 + b^2}{a^2 - b^2} \) and determine the conditions under which \( \theta \) takes real values.

### Step-by-Step Solution:

1. **Understanding Secant Function**:
\[ \sec \theta = \frac{1}{\cos \theta} \]
Therefore, we can express \( \cos \theta \) as:
\[ \cos \theta = \frac{a^2 - b^2}{a^2 + b^2} \]

2. **Condition for Cosine**:
The cosine function must satisfy the condition:
\[ -1 \leq \cos \theta \leq 1 \]
This leads us to two inequalities:
\[ -1 \leq \frac{a^2 - b^2}{a^2 + b^2} \leq 1 \]

3. **Analyzing the Right Inequality**:
Starting with the right inequality:
\[ \frac{a^2 - b^2}{a^2 + b^2} \leq 1 \]
This simplifies to:
\[ a^2 - b^2 \leq a^2 + b^2 \]
Rearranging gives:
\[ -b^2 \leq b^2 \implies 0 \leq 2b^2 \implies b^2 \geq 0 \]
Since \( b^2 \) is always non-negative, this condition is satisfied for all real \( b \).

4. **Analyzing the Left Inequality**:
Now, consider the left inequality:
\[ -1 \leq \frac{a^2 - b^2}{a^2 + b^2} \]
This simplifies to:
\[ - (a^2 + b^2) \leq a^2 - b^2 \]
Rearranging gives:
\[ 0 \leq 2a^2 \implies a^2 \geq 0 \]
This condition is also satisfied for all real \( a \).

5. **Condition for Non-zero Denominator**:
We must also ensure that the denominator \( a^2 + b^2 \neq 0 \) and \( a^2 - b^2 \neq 0 \):
- \( a^2 + b^2 = 0 \) implies \( a = 0 \) and \( b = 0 \), which is not allowed.
- \( a^2 - b^2 = 0 \) implies \( a^2 = b^2 \), leading to \( |a| = |b| \).

6. **Final Conditions**:
Therefore, for \( \sec \theta \) to yield real values of \( \theta \), we conclude:
- \( a^2 \neq b^2 \) (or \( |a| \neq |b| \))
- \( a \neq 0 \) and \( b \neq 0 \)

### Conclusion:
The values of \( \theta \) are real if and only if \( |a| \neq |b| \) and \( a \neq 0 \), \( b \neq 0 \).