`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 condition under which the expression for sec θ gives real values of θ. The expression given is:

\[ \sec \theta = \frac{a^2 + b^2}{a^2 - b^2} \]

### Step 1: Convert secant to cosine
We know that sec θ is the reciprocal of cos θ. Therefore, we can write:

\[ \cos \theta = \frac{a^2 - b^2}{a^2 + b^2} \]

### Step 2: Identify the condition for cos θ
The cosine function has a range of values between -1 and 1. Hence, we need to ensure that:

\[ -1 \leq \frac{a^2 - b^2}{a^2 + b^2} \leq 1 \]

### Step 3: Analyze the inequality
We will break this down into two inequalities.

1. **Upper Bound**:
\[ \frac{a^2 - b^2}{a^2 + b^2} \leq 1 \]
This implies:
\[ a^2 - b^2 \leq a^2 + b^2 \]
Simplifying this gives:
\[ -b^2 \leq b^2 \implies 0 \leq 2b^2 \implies b^2 \geq 0 \]
This is always true for real numbers.

2. **Lower Bound**:
\[ \frac{a^2 - b^2}{a^2 + b^2} \geq -1 \]
This implies:
\[ a^2 - b^2 \geq - (a^2 + b^2) \]
Simplifying this gives:
\[ a^2 - b^2 + a^2 + b^2 \geq 0 \implies 2a^2 \geq 0 \implies a^2 \geq 0 \]
This is also always true for real numbers.

### Step 4: Condition for a^2 - b^2
We also need to ensure that the denominator \( a^2 - b^2 \) is not zero, which gives us the condition:

\[ a^2 - b^2 \neq 0 \implies a^2 \neq b^2 \implies a \neq \pm b \]

### Step 5: Conclusion
From the analysis, we conclude that for sec θ to yield real values of θ, the conditions are:

1. \( a^2 \neq b^2 \) (i.e., \( a \neq b \) and \( a \neq -b \))
2. \( a \) and \( b \) can be any real numbers, but they must not be equal in absolute value.

Thus, the final condition for real values of θ is:

\[ |a| \neq |b| \]