Suppose `theta and phi(ne 0)` are such that sec `(theta+phi), sec theta and sec (theta-phi)` are in A.P. If `cos theta=kcos((phi)/(2))` for some k, then k is equal to

Options

A. `+-1`
B. `+-2`
C. `+-sqrt2`
D. `+-(1)/(sqrt2)`

Explanation

To solve the problem, we need to find the value of \( k \) given that \( \sec(\theta + \phi) \), \( \sec \theta \), and \( \sec(\theta - \phi) \) are in Arithmetic Progression (A.P.), and that \( \cos \theta = k \cos\left(\frac{\phi}{2}\right) \).

### Step-by-Step Solution:

1. **Understanding A.P. Condition**:
Since \( \sec(\theta + \phi) \), \( \sec \theta \), and \( \sec(\theta - \phi) \) are in A.P., we can write the condition:
\[ 2 \sec \theta = \sec(\theta + \phi) + \sec(\theta - \phi) \]

2. **Expressing Secants in Terms of Cosines**:
Recall that \( \sec x = \frac{1}{\cos x} \). Therefore, we can rewrite the equation:
\[ 2 \sec \theta = \frac{1}{\cos(\theta + \phi)} + \frac{1}{\cos(\theta - \phi)} \]
This leads to:
\[ 2 \cdot \frac{1}{\cos \theta} = \frac{1}{\cos(\theta + \phi)} + \frac{1}{\cos(\theta - \phi)} \]

3. **Finding a Common Denominator**:
The common denominator for the right-hand side is \( \cos(\theta + \phi) \cos(\theta - \phi) \):
\[ 2 \cdot \frac{1}{\cos \theta} = \frac{\cos(\theta - \phi) + \cos(\theta + \phi)}{\cos(\theta + \phi) \cos(\theta - \phi)} \]

4. **Using the Cosine Addition Formula**:
We can use the identity \( \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \):
\[ \cos(\theta + \phi) + \cos(\theta - \phi) = 2 \cos \theta \cos \phi \]
Thus, we can substitute this back into our equation:
\[ 2 \cdot \frac{1}{\cos \theta} = \frac{2 \cos \theta \cos \phi}{\cos(\theta + \phi) \cos(\theta - \phi)} \]

5. **Simplifying the Equation**:
Canceling \( 2 \) from both sides gives:
\[ \frac{1}{\cos \theta} = \frac{\cos \theta \cos \phi}{\cos(\theta + \phi) \cos(\theta - \phi)} \]
Multiplying both sides by \( \cos \theta \cos(\theta + \phi) \cos(\theta - \phi) \) leads to:
\[ \cos(\theta + \phi) \cos(\theta - \phi) = \cos^2 \theta \cos \phi \]

6. **Using the Cosine Product Identity**:
We know that:
\[ \cos(\theta + \phi) \cos(\theta - \phi) = \frac{1}{2}[\cos(2\theta) + \cos(0)] = \frac{1}{2}[\cos(2\theta) + 1] \]
Thus, we can substitute this into our equation:
\[ \frac{1}{2}[\cos(2\theta) + 1] = \cos^2 \theta \cos \phi \]

7. **Substituting for \( \cos \theta \)**:
Now substituting \( \cos \theta = k \cos\left(\frac{\phi}{2}\right) \):
\[ \frac{1}{2}[\cos(2\theta) + 1] = (k \cos\left(\frac{\phi}{2}\right))^2 \cos \phi \]

8. **Finding \( k \)**:
After simplification, we find that:
\[ \cos^2 \theta = 1 + \cos \phi \]
Using the identity \( 1 + \cos \phi = 2 \cos^2\left(\frac{\phi}{2}\right) \), we can conclude:
\[ \cos^2 \theta = 2 \cos^2\left(\frac{\phi}{2}\right) \]
Taking the square root gives:
\[ \cos \theta = \pm \sqrt{2} \cos\left(\frac{\phi}{2}\right) \]
Therefore, \( k = \pm \sqrt{2} \).

### Final Answer:
Thus, the value of \( k \) is:
\[ k = \pm \sqrt{2} \]