If `cos(theta-alpha),costheta,cos(theta+alpha)` are in H.P., then `costheta sec""(alpha)/(2)=?`

Options

A. 0
B. `pm1`
C. 2
D. `pmsqrt(2)`

Answer

To solve the problem where \( \cos(\theta - \alpha), \cos(\theta), \cos(\theta + \alpha) \) are in Harmonic Progression (H.P.), we need to find the value of \( \frac{\cos(\theta) \sec(\alpha)}{2} \).

### Step-by-Step Solution:

1. **Understanding Harmonic Progression**:
If three numbers \( a, b, c \) are in H.P., then the relationship can be expressed as:
\[ \frac{1}{b} = \frac{2}{a + c} \]
Here, let \( a = \cos(\theta - \alpha) \), \( b = \cos(\theta) \), and \( c = \cos(\theta + \alpha) \).

2. **Setting up the equation**:
From the H.P. condition, we have:
\[ \frac{1}{\cos(\theta)} = \frac{2}{\cos(\theta - \alpha) + \cos(\theta + \alpha)} \]

3. **Using the cosine addition formulas**:
We know that:
\[ \cos(\theta - \alpha) + \cos(\theta + \alpha) = 2 \cos(\theta) \cos(\alpha) \]
Substituting this into the equation gives:
\[ \frac{1}{\cos(\theta)} = \frac{2}{2 \cos(\theta) \cos(\alpha)} \]

4. **Simplifying the equation**:
This simplifies to:
\[ \frac{1}{\cos(\theta)} = \frac{1}{\cos(\theta) \cos(\alpha)} \]
Multiplying both sides by \( \cos(\theta) \) (assuming \( \cos(\theta) \neq 0 \)):
\[ 1 = \cos(\alpha) \]

5. **Finding the value of \( \frac{\cos(\theta) \sec(\alpha)}{2} \)**:
Since \( \cos(\alpha) = 1 \), it follows that \( \sec(\alpha) = \frac{1}{\cos(\alpha)} = 1 \). Therefore:
\[ \frac{\cos(\theta) \sec(\alpha)}{2} = \frac{\cos(\theta) \cdot 1}{2} = \frac{\cos(\theta)}{2} \]

6. **Conclusion**:
Thus, the final answer is:
\[ \frac{\cos(\theta)}{2} \]