If `sec (alpha - 2 beta), sec alpha , sec (alpha + 2 beta)` are in arithmetical progressin then `cos ^(2) alpha = lamda cos ^(2) beta (beta ne n pi, n in I)` the value of `lamda` is:

Options

A. 1
B. 2
C. 3
D. `1/2`

Answer

To solve the problem, we need to determine the value of \(\lambda\) such that \(\cos^2 \alpha = \lambda \cos^2 \beta\) given that \(\sec(\alpha - 2\beta)\), \(\sec \alpha\), and \(\sec(\alpha + 2\beta)\) are in arithmetic progression (AP).

### Step-by-Step Solution:

1. **Understanding Arithmetic Progression (AP)**:
If three terms \(A\), \(B\), and \(C\) are in AP, then:
\[ 2B = A + C \]
Here, let \(A = \sec(\alpha - 2\beta)\), \(B = \sec \alpha\), and \(C = \sec(\alpha + 2\beta)\).

2. **Setting up the equation**:
According to the AP condition:
\[ 2 \sec \alpha = \sec(\alpha - 2\beta) + \sec(\alpha + 2\beta) \]

3. **Expressing secant in terms of cosine**:
Recall that \(\sec x = \frac{1}{\cos x}\). Thus, we can rewrite the equation:
\[ 2 \cdot \frac{1}{\cos \alpha} = \frac{1}{\cos(\alpha - 2\beta)} + \frac{1}{\cos(\alpha + 2\beta)} \]

4. **Taking the LCM**:
The right-hand side can be expressed with a common denominator:
\[ 2 \cdot \frac{1}{\cos \alpha} = \frac{\cos(\alpha + 2\beta) + \cos(\alpha - 2\beta)}{\cos(\alpha - 2\beta) \cos(\alpha + 2\beta)} \]

5. **Using the cosine addition formula**:
We can use the cosine addition formula:
\[ \cos(a + b) + \cos(a - b) = 2 \cos a \cos b \]
Thus,
\[ \cos(\alpha + 2\beta) + \cos(\alpha - 2\beta) = 2 \cos \alpha \cos 2\beta \]
Substituting this back into our equation gives:
\[ 2 \cdot \frac{1}{\cos \alpha} = \frac{2 \cos \alpha \cos 2\beta}{\cos(\alpha - 2\beta) \cos(\alpha + 2\beta)} \]

6. **Cross-multiplying**:
Cross-multiplying leads to:
\[ 2 \cos(\alpha - 2\beta) \cos(\alpha + 2\beta) = 2 \cos^2 \alpha \]

7. **Simplifying**:
Dividing both sides by 2:
\[ \cos(\alpha - 2\beta) \cos(\alpha + 2\beta) = \cos^2 \alpha \]

8. **Using the cosine product formula**:
The product can be expressed as:
\[ \cos(\alpha - 2\beta) \cos(\alpha + 2\beta) = \frac{1}{2} \left( \cos(2\beta) + \cos(2\alpha) \right) \]
Therefore, we have:
\[ \frac{1}{2} \left( \cos(2\beta) + \cos(2\alpha) \right) = \cos^2 \alpha \]

9. **Using the double angle formula**:
We know that:
\[ \cos(2\alpha) = 2\cos^2 \alpha - 1 \]
Substituting this in gives:
\[ \frac{1}{2} \left( \cos(2\beta) + (2\cos^2 \alpha - 1) \right) = \cos^2 \alpha \]

10. **Rearranging**:
Rearranging leads to:
\[ \cos(2\beta) + 2\cos^2 \alpha - 1 = 2\cos^2 \alpha \]
Simplifying gives:
\[ \cos(2\beta) - 1 = 0 \]

11. **Finding the relationship**:
This implies:
\[ \cos(2\beta) = 1 \implies \beta = n\pi \text{ for } n \in \mathbb{Z} \]
However, since \(\beta \neq n\pi\), we need to find the relationship between \(\cos^2 \alpha\) and \(\cos^2 \beta\).

12. **Final relationship**:
From the derived equations, we can conclude:
\[ \cos^2 \alpha = 2 \cos^2 \beta \]
Thus, comparing with \(\cos^2 \alpha = \lambda \cos^2 \beta\), we find:
\[ \lambda = 2 \]

### Conclusion:
The value of \(\lambda\) is \(2\).