Here,
$\frac{1}{\sin 1^{\circ} \sin 2^{\circ}}+\frac{1}{\sin 2^{\circ} \sin 3^{\circ}}+\ldots+\frac{1}{\sin 89^{\circ} \sin 90^{\circ}}$
$=\frac{1}{\sin 1^{\circ}}\left[\begin{array}{l}\frac{\sin \left(2^{\circ}-1^{\circ}\right)}{\sin 1^{\circ} \sin 2^{\circ}}+\frac{\sin \left(3^{\circ}-2^{\circ}\right)}{\sin 2^{\circ} \sin 3^{\circ}}+\ldots . . \\ +\frac{\sin \left(90^{\circ}-89^{\circ}\right)}{\sin 89^{\circ} \sin 90^{\circ}}\end{array}\right]$
We know that
$\sin (A-B)=\sin A \cdot \cos B-\cos A \cdot \sin B$
$\begin{aligned}=\frac{1}{\sin l^{\circ}}\left[\begin{array}{l}\frac{\sin 2^{\circ} \cdot \cos 1^{\circ}-\cos 2^{\circ} \cdot \sin l^{\circ}}{\sin 1^{\circ} \cdot \sin 2^{\circ}} \\ +\frac{\sin 3^{\circ} \cdot \cos 2^{\circ}-\cos 3^{\circ} \cdot \sin 2^{\circ}}{\sin 3^{\circ} \cdot \sin 2^{\circ}}+\ldots . \\ \quad \ldots .+\frac{\sin 90^{\circ} \cdot \cos 89^{\circ}-\cos 90^{\circ} \cdot \sin 89^{\circ}}{\sin 89^{\circ} \cdot \sin 90^{\circ}}\end{array}\right]\end{aligned}$
$\begin{array}{r}=\frac{1}{\sin 1^{\circ}}\left[\cot 1^{\circ}-\cot 2^{\circ}+\cot 2^{\circ}-\cot 3^{\circ}+\ldots .\right. \\ \left.\quad+\cot 89^{\circ}-\cot 90^{\circ}\right]\end{array}$
$=\frac{1}{\sin 1^{\circ}}\left[\cot 1^{\circ}-\cot 90^{\circ}\right]=\frac{\cot 1^{\circ}}{\sin 1^{\circ}}=\frac{\cos 1^{\circ}}{\sin ^2 1^{\circ}}$