AIME Problem

Note first that for the terms from $\dfrac{1}{\sin 91^{\circ} \sin 92^{\circ}}$ onward, we can rewrite them as

$\dfrac{1}{\sin(180^{\circ} - k^{\circ}) \sin(180^{\circ} - (k + 1)^{\circ})}$ .

These terms, in reverse order, can then be written as

$\dfrac{1}{\sin 46^{\circ} \sin 47^{\circ}} + \dfrac{1}{\sin 48^{\circ} \sin 49^{\circ}} + .... + \dfrac{1}{\sin 88^{\circ} \sin 89^{\circ}}$ .

The entirety of the given sum can then be written as $\displaystyle\sum_{k=45}^{89} \dfrac{1}{\sin k^{\circ} \sin (k + 1)^{\circ}}$ .

Now anticipating a telescoping series on the horizon, note that

$\dfrac{1}{\sin k^{\circ} \sin (k + 1)^{\circ}} = \dfrac{\sin ((k + 1) - k)^{\circ}}{\sin 1^{\circ}} \times \dfrac{1}{\sin k^{\circ} \sin(k + 1)^{\circ}} =$

$\dfrac{1}{\sin 1^{\circ}} \times \dfrac{\sin(k + 1)^{\circ} \cos k^{\circ} - \cos(k + 1)^{\circ} \sin k^{\circ}}{\sin k^{\circ} \sin(k + 1)^{\circ}} = \dfrac{1}{\sin 1^{\circ}} \times (\cot k^{\circ} - \cot(k + 1)^{\circ})$ .

The given sum then telescopes to

$\dfrac{1}{\sin 1^{\circ}} \displaystyle\sum_{k=45}^{89} (\cot k^{\circ} - \cot(k + 1)^{\circ}) = \dfrac{1}{\sin 1^{\circ}} \times (\cot 45^{\circ} - \cot 90^{\circ}) = \dfrac{1}{\sin 1^{\circ}} \times (1 - 0) = \dfrac{1}{\sin 1^{\circ}}$ .

The correct option is thus $n = \boxed{1}$ .