`alpha,beta and gamma` are parametric angles of three points P, Q and R respectively, on the circle `x^2 + y^2 = 1` and A is the point (-1, 0). If the lengths of the chords AP, AQ and AR are in GP, then `cos alpha/2, cos beta/2 and cos gamma/2` are in

Options

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

Answer

To solve the problem, we need to find the relationship between the cosines of half the angles corresponding to points P, Q, and R on the circle, given that the lengths of the chords from point A to these points are in geometric progression (GP).

### Step-by-step Solution:

1. **Identify the Points on the Circle**:
The points P, Q, and R on the unit circle can be represented in parametric form as:
- \( P(\cos \alpha, \sin \alpha) \)
- \( Q(\cos \beta, \sin \beta) \)
- \( R(\cos \gamma, \sin \gamma) \)

Here, \( \alpha, \beta, \gamma \) are the angles corresponding to points P, Q, and R respectively.

2. **Coordinates of Point A**:
The coordinates of point A are given as \( A(-1, 0) \).

3. **Calculate Lengths of Chords**:
The lengths of the chords \( AP, AQ, AR \) can be calculated using the distance formula:
- Length of chord \( AP \):
\[ AP = \sqrt{(\cos \alpha + 1)^2 + (\sin \alpha - 0)^2} = \sqrt{(\cos \alpha + 1)^2 + \sin^2 \alpha} \]
Expanding this:
\[ AP = \sqrt{(\cos^2 \alpha + 2\cos \alpha + 1 + \sin^2 \alpha)} = \sqrt{2 + 2\cos \alpha} = \sqrt{2(1 + \cos \alpha)} = \sqrt{2} \sqrt{1 + \cos \alpha} \]

- Length of chord \( AQ \):
\[ AQ = \sqrt{(\cos \beta + 1)^2 + (\sin \beta - 0)^2} = \sqrt{2(1 + \cos \beta)} = \sqrt{2} \sqrt{1 + \cos \beta} \]

- Length of chord \( AR \):
\[ AR = \sqrt{(\cos \gamma + 1)^2 + (\sin \gamma - 0)^2} = \sqrt{2(1 + \cos \gamma)} = \sqrt{2} \sqrt{1 + \cos \gamma} \]

4. **Condition of Geometric Progression**:
Since the lengths \( AP, AQ, AR \) are in GP, we have:
\[ AQ^2 = AP \cdot AR \]
Substituting the lengths:
\[ ( \sqrt{2} \sqrt{1 + \cos \beta} )^2 = ( \sqrt{2} \sqrt{1 + \cos \alpha} ) \cdot ( \sqrt{2} \sqrt{1 + \cos \gamma} ) \]
Simplifying gives:
\[ 2(1 + \cos \beta) = 2\sqrt{(1 + \cos \alpha)(1 + \cos \gamma)} \]
Dividing both sides by 2:
\[ 1 + \cos \beta = \sqrt{(1 + \cos \alpha)(1 + \cos \gamma)} \]

5. **Using Cosine Half-Angle Identity**:
We can use the identity \( 1 + \cos x = 2 \cos^2 \frac{x}{2} \):
- Thus, we have:
\[ 1 + \cos \alpha = 2 \cos^2 \frac{\alpha}{2}, \quad 1 + \cos \beta = 2 \cos^2 \frac{\beta}{2}, \quad 1 + \cos \gamma = 2 \cos^2 \frac{\gamma}{2} \]
Substituting these into the equation:
\[ 2 \cos^2 \frac{\beta}{2} = \sqrt{(2 \cos^2 \frac{\alpha}{2})(2 \cos^2 \frac{\gamma}{2})} \]
Simplifying gives:
\[ \cos^2 \frac{\beta}{2} = \cos \frac{\alpha}{2} \cdot \cos \frac{\gamma}{2} \]

6. **Conclusion**:
This shows that \( \cos \frac{\alpha}{2}, \cos \frac{\beta}{2}, \cos \frac{\gamma}{2} \) are in geometric progression (GP).

### Final Answer:
Thus, \( \cos \frac{\alpha}{2}, \cos \frac{\beta}{2}, \cos \frac{\gamma}{2} \) are in GP.