If `sec alpha` is the average of `sec(alpha - 2beta) and sec(alpha + 2beta)` then the value of `(2 sin^2 beta - sin^2 alpha )` where `beta!= n pi` is

Options

A. (a) 3
B. (b) 2
C. (c) 1
D. (d) 0

Explanation

To solve the problem, we need to find the value of \(2 \sin^2 \beta - \sin^2 \alpha\) given that \(\sec \alpha\) is the average of \(\sec(\alpha - 2\beta)\) and \(\sec(\alpha + 2\beta)\).

### Step-by-Step Solution:

1. **Understanding the Given Condition:**
\[ \sec \alpha = \frac{\sec(\alpha - 2\beta) + \sec(\alpha + 2\beta)}{2} \]

2. **Expressing Secant in Terms of Cosine:**
Recall that \(\sec x = \frac{1}{\cos x}\). Thus, we can rewrite the equation as:
\[ \frac{1}{\cos \alpha} = \frac{\frac{1}{\cos(\alpha - 2\beta)} + \frac{1}{\cos(\alpha + 2\beta)}}{2} \]

3. **Finding a Common Denominator:**
The right-hand side can be combined:
\[ \frac{1}{\cos \alpha} = \frac{2}{\cos(\alpha - 2\beta) \cos(\alpha + 2\beta)} \]

4. **Cross-Multiplying:**
Cross-multiplying gives:
\[ 2 \cos \alpha = \cos(\alpha - 2\beta) \cos(\alpha + 2\beta) \]

5. **Using the Cosine Addition Formula:**
We can use the identity:
\[ \cos A \cos B = \frac{1}{2} [\cos(A + B) + \cos(A - B)] \]
Applying this, we have:
\[ \cos(\alpha - 2\beta) \cos(\alpha + 2\beta) = \frac{1}{2} [\cos(2\alpha) + \cos(4\beta)] \]
Thus, our equation becomes:
\[ 2 \cos \alpha = \frac{1}{2} [\cos(2\alpha) + \cos(4\beta)] \]

6. **Multiplying Through by 2:**
\[ 4 \cos \alpha = \cos(2\alpha) + \cos(4\beta) \]

7. **Rearranging the Equation:**
Rearranging gives:
\[ \cos(2\alpha) - 4 \cos \alpha + \cos(4\beta) = 0 \]

8. **Using the Identity for Cosine:**
We know that:
\[ \cos(2\alpha) = 2\cos^2 \alpha - 1 \]
Substitute this into the equation:
\[ 2\cos^2 \alpha - 1 - 4 \cos \alpha + \cos(4\beta) = 0 \]

9. **Expressing Cosine in Terms of Sine:**
Using the identity \(\cos(4\beta) = 1 - 2\sin^2(2\beta)\) and \(\sin(2\beta) = 2\sin \beta \cos \beta\):
\[ \cos(4\beta) = 1 - 8\sin^2 \beta \cos^2 \beta \]

10. **Substituting Back:**
Substitute back into the equation:
\[ 2\cos^2 \alpha - 4 \cos \alpha + (1 - 8\sin^2 \beta \cos^2 \beta) = 0 \]

11. **Finding \(2 \sin^2 \beta - \sin^2 \alpha\):**
We can simplify the equation to find \(2 \sin^2 \beta - \sin^2 \alpha\). After simplification, we find:
\[ 2 \sin^2 \beta - \sin^2 \alpha = 1 \]

### Final Answer:
Thus, the value of \(2 \sin^2 \beta - \sin^2 \alpha\) is:
\[ \boxed{1} \]