`alpha,beta,gammaand delta` are angles in I,II,II and IV quadrants, respectively and none of them is an integral multiple of `pi//2`. They form an increasing arithmetic progression.
Which of the following holds?

Options

A. `cos(alpha-delta)gt0`
B. `cos(alpha-delta)=0`
C. `cos(alpha-delta)lt0`
D. `cos(alpha-delta)gt0or cos(alpha-delta)lt0`

Explanation

To solve the problem, we need to analyze the angles α, β, γ, and δ, which are in the first, second, third, and fourth quadrants respectively, and they form an increasing arithmetic progression (AP).

### Step-by-Step Solution

1. **Understanding Quadrants**:
- α is in the first quadrant: \( 0 < α < \frac{\pi}{2} \)
- β is in the second quadrant: \( \frac{\pi}{2} < β < \pi \)
- γ is in the third quadrant: \( \pi < γ < \frac{3\pi}{2} \)
- δ is in the fourth quadrant: \( \frac{3\pi}{2} < δ < 2\pi \)

2. **Arithmetic Progression**:
Since α, β, γ, and δ are in an increasing AP, we can express them in terms of a common difference \( d \):
- β = α + d
- γ = α + 2d
- δ = α + 3d

3. **Establishing Inequalities**:
From the definitions of the quadrants, we can establish the following inequalities:
- For β: \( \frac{\pi}{2} < α + d < \pi \)
- For γ: \( \pi < α + 2d < \frac{3\pi}{2} \)
- For δ: \( \frac{3\pi}{2} < α + 3d < 2\pi \)

4. **Finding Relationships**:
From the inequality for β:
\[ d > \frac{\pi}{2} - α \quad \text{(1)} \]
From the inequality for γ:
\[ 2d > \pi - α \quad \Rightarrow \quad d > \frac{\pi - α}{2} \quad \text{(2)} \]
From the inequality for δ:
\[ 3d > \frac{3\pi}{2} - α \quad \Rightarrow \quad d > \frac{\frac{3\pi}{2} - α}{3} \quad \text{(3)} \]

5. **Combining Inequalities**:
Now we need to analyze these inequalities to find a consistent value for \( d \). Since \( d \) must satisfy all three inequalities, we can set up a relationship:
- From (1) and (2):
\[ \frac{\pi}{2} - α < \frac{\pi - α}{2} \quad \Rightarrow \quad \pi - 2α < \frac{\pi}{2} \quad \Rightarrow \quad 2α > \frac{\pi}{2} \quad \Rightarrow \quad α > \frac{\pi}{4} \]

6. **Conclusion**:
Since α is in the first quadrant, and we've established that \( α > \frac{\pi}{4} \), we can conclude that:
- \( \cos(α) > 0 \)
- \( \cos(β) > 0 \)
- \( \cos(γ) < 0 \)
- \( \cos(δ) > 0 \)

Therefore, the cosine of the differences:
- \( \cos(α - γ) \) and \( \cos(α - δ) \) will both be greater than 0, confirming that the correct option is that they are greater than or equal to 0.

### Final Answer
The correct statement is that \( \cos(α - γ) > 0 \) and \( \cos(α - δ) > 0 \).