Step-by-step Solution:

If \( a_1, a_2, \ldots, a_n \) are in arithmetic progression (AP), then: \[ a_2 - a_1 = a_3 - a_2 = \ldots = a_n - a_{n-1} = d, \] where \( d \) is the common difference. The expression: \[ \sin d \left( \frac{1}{\sin a_1 \cdot \sin a_2} + \frac{1}{\sin a_2 \cdot \sin a_3} + \ldots + \frac{1}{\sin a_{n-1} \cdot \sin a_n} \right) \] is simplified as: \[ \frac{\sin d}{\sin a_1 \cdot \sin a_2} + \frac{\sin d}{\sin a_2 \cdot \sin a_3} + \ldots + \frac{\sin d}{\sin a_{n-1} \cdot \sin a_n}. \] Using the trigonometric identity: \[ \sin(x-y) = \sin x \cos y - \cos x \sin y, \] each term simplifies to: \[ \frac{\sin a_2 \cos a_1 - \cos a_2 \sin a_1}{\sin a_1 \cdot \sin a_2} + \frac{\sin a_3 \cos a_2 - \cos a_3 \sin a_2}{\sin a_2 \cdot \sin a_3} + \ldots. \] This telescopes to: \[ \cot a_1 - \cot a_n. \] Hence: \[ \sin d \sum_{k=1}^{n-1} \frac{1}{\sin a_k \cdot \sin a_{k+1}} = \cot a_1 - \cot a_n. \]