Given,
$$
\begin{aligned}
\frac{1}{\sin 45^{\circ} \sin 46^{\circ}}+\frac{1}{\sin 46^{\circ} \sin 47^{\circ}}+\ldots \ldots .+\frac{1}{\sin 89^{\circ} \sin 90^{\circ}} \\
=\frac{1}{\sin 1}\left[\frac{\sin 1}{\sin 45^{\circ} \sin 46^{\circ}}+\frac{\sin 1}{\sin 46^{\circ} \sin 47^{\circ}}+\ldots . .+\frac{\sin 1}{\sin 89^{\circ} \sin 90^{\circ}}\right]
\end{aligned}
$$
$$
\begin{gathered}
=\frac{1}{\sin 1}\left[\frac{\sin \left(46^{\circ}-45^{\circ}\right)}{\sin 45^{\circ} \sin 46^{\circ}}+\frac{\sin \left(47^{\circ}-46^{\circ}\right)}{\sin 46^{\circ} \sin 47^{\circ}}+\ldots .+\frac{\sin \left(90^{\circ}-89^{\circ}\right)}{\sin 89^{\circ} \sin 90^{\circ}}\right] \\
=\frac{1}{\sin 1}\left[\frac{\sin 46^{\circ} \cos 45^{\circ}-\cos 46^{\circ} \sin 45^{\circ}}{\sin 45^{\circ} \sin 46^{\circ}}+\ldots . .+ \text { same up to } 45 \text { terms }\right] \\
=\frac{1}{\sin 1}\left[\cot 45^{\circ}-\cot 46^{\circ}+\cot 46^{\circ}-\cot 47^{\circ}+\ldots+\cot 90^{\circ}-\cot 89^{\circ}\right] \\
\quad=\frac{1}{\sin 1}\left[\cot 45^{\circ}-\cot 90^{\circ}\right]=\frac{1}{\sin 1}[1-0]=\frac{1}{\sin 1}
\end{gathered}
$$
Now, $\frac{1}{\sin 1}=\frac{1}{\sin x}$
So, $x=1$
Take $\sin \left(\frac{\pi}{2} x\right)=\sin \left(\frac{\pi}{2}\right)=1$
So, option (c) is correct.