**Given:**
- A ladder is resting against a vertical wall.
- The angle between the ladder and the horizontal is α.
- The foot of the ladder is pulled away from the wall through a distance p.
- The upper end of the ladder slides a distance q along the wall.
- The angle between the ladder and the horizontal becomes β.

**To prove:**
p/q = (cosβ - cosα)/(sinα - sinβ)

**Explanation:**

1. Let's consider a right-angled triangle formed by the ladder, the wall, and the ground. The ladder acts as the hypotenuse of the triangle.
2. The length of the ladder can be represented as L.
3. The distance of the foot of the ladder from the wall is p, and the distance of the upper end of the ladder from the wall is q.
4. The height of the wall can be represented as h.
5. The angle α is the angle between the ladder and the horizontal, and the angle β is the angle between the ladder and the ground.
6. Using trigonometry, we can write the following equations:
- cosα = p/L (1)
- cosβ = q/L (2)
- sinα = h/L (3)
- sinβ = h/(L+q) (4)
7. We need to find the value of p/q.
8. Rearranging equation (1), we get L = p/cosα.
9. Substituting the value of L in equation (2), we get cosβ = q/(p/cosα) = qcosα/p.
10. Rearranging equation (3), we get h = Lsinα.
11. Substituting the value of L in equation (3), we get h = (p/cosα)sinα = psinα/cosα = ptanα.
12. Rearranging equation (4), we get sinβ = h/(L+q) = (ptanα)/(p/cosα+q).
13. Multiplying both sides of equation (4) by (p/cosα+q), we get sinβ(p/cosα+q) = ptanα.
14. Expanding and rearranging the equation, we get p(sinβcosα+qsinβ) = ptanαcosα+qsinβtanα.
15. Dividing both sides of the equation by p, we get sinβcosα+qsinβ = tanαcosα+qsinβtanα/p.
16. Rearranging the equation, we get qsinβ - qsinβtanα/p = cosα - cosαtanα.
17. Dividing both sides of the equation by sinα-sinβ, we get (qsinβ - qsinβtanα/p)/(sinα - sinβ) = (cosα - cosαtanα)/(sinα - sinβ).
18. Simplifying the equation, we get p/q = (cosβ - cosα)/(sinα - sinβ).

Hence, we have proved that p/q = (cosβ - cosα)/(sinα - sinβ).