If `cos^4 theta sec^2 alpha,1/2` and `sin^4 theta cosec^2 alpha` are in `A.P.`, then `cos^8 theta sec^6 alpha ,1/2` and `sin^8 theta cosec^6 alpha` are

Options

A. AP
B. GP
C. HP
D. None of these

Answer

To solve the problem, we need to show that if \( \cos^4 \theta \sec^2 \alpha, \frac{1}{2}, \sin^4 \theta \csc^2 \alpha \) are in Arithmetic Progression (A.P.), then \( \cos^8 \theta \sec^6 \alpha, \frac{1}{2}, \sin^8 \theta \csc^6 \alpha \) are also in A.P.

### Step-by-step Solution:

1. **Understanding the A.P. Condition**:
For three numbers \( A, B, C \) to be in A.P., the condition is:
\[ 2B = A + C \]
Here, let:
- \( A = \cos^4 \theta \sec^2 \alpha \)
- \( B = \frac{1}{2} \)
- \( C = \sin^4 \theta \csc^2 \alpha \)

Thus, we have:
\[ 2 \cdot \frac{1}{2} = \cos^4 \theta \sec^2 \alpha + \sin^4 \theta \csc^2 \alpha \]
Simplifying gives:
\[ 1 = \cos^4 \theta \sec^2 \alpha + \sin^4 \theta \csc^2 \alpha \]

2. **Expressing in Terms of Sine and Cosine**:
We know:
\[ \sec^2 \alpha = \frac{1}{\cos^2 \alpha} \quad \text{and} \quad \csc^2 \alpha = \frac{1}{\sin^2 \alpha} \]
Therefore, we can rewrite the equation:
\[ 1 = \cos^4 \theta \cdot \frac{1}{\cos^2 \alpha} + \sin^4 \theta \cdot \frac{1}{\sin^2 \alpha} \]
This simplifies to:
\[ 1 = \frac{\cos^4 \theta}{\cos^2 \alpha} + \frac{\sin^4 \theta}{\sin^2 \alpha} \]

3. **Cross Multiplying**:
Multiply through by \( \cos^2 \alpha \sin^2 \alpha \):
\[ \cos^4 \theta \sin^2 \alpha + \sin^4 \theta \cos^2 \alpha = \cos^2 \alpha \sin^2 \alpha \]

4. **Rearranging**:
Rearranging gives:
\[ \cos^4 \theta \sin^2 \alpha + \sin^4 \theta \cos^2 \alpha - \cos^2 \alpha \sin^2 \alpha = 0 \]

5. **Factoring**:
This can be factored as:
\[ (\cos^2 \theta \sin^2 \alpha - \cos^2 \alpha)(\cos^2 \theta \sin^2 \alpha + \sin^2 \theta) = 0 \]
This implies:
\[ \cos^2 \theta = \cos^2 \alpha \]

6. **Applying the Result**:
Now we need to check if \( \cos^8 \theta \sec^6 \alpha, \frac{1}{2}, \sin^8 \theta \csc^6 \alpha \) are in A.P. Using the same substitution:
\[ A' = \cos^8 \theta \sec^6 \alpha, \quad B' = \frac{1}{2}, \quad C' = \sin^8 \theta \csc^6 \alpha \]
We can express:
\[ A' = \frac{\cos^8 \theta}{\cos^6 \alpha}, \quad C' = \frac{\sin^8 \theta}{\sin^6 \alpha} \]

7. **Using the Previous Result**:
Since \( \cos^2 \theta = \cos^2 \alpha \) and \( \sin^2 \theta = \sin^2 \alpha \), we can substitute:
\[ A' = \frac{\cos^8 \alpha}{\cos^6 \alpha} = \cos^2 \alpha \]
\[ C' = \frac{\sin^8 \alpha}{\sin^6 \alpha} = \sin^2 \alpha \]

8. **Final A.P. Condition**:
Now, we check:
\[ 2B' = A' + C' \]
\[ 1 = \cos^2 \alpha + \sin^2 \alpha \]
This is true, thus confirming that:
\[ \cos^8 \theta \sec^6 \alpha, \frac{1}{2}, \sin^8 \theta \csc^6 \alpha \text{ are in A.P.} \]