Correct Answer: $1$

Solution : Given: $\frac{\sin 30^{\circ} \sin 40^{\circ} \sin 50^{\circ} \sin 60^{\circ}}{\cos 30^{\circ} \cos 40^{\circ} \cos 50^{\circ} \cos 60^{\circ}}$
= $\frac{\sin (90^{\circ}-60^{\circ}) \sin (90^{\circ}-50^{\circ}) \sin (90^{\circ}-40^{\circ}) \sin (90^{\circ}-30^{\circ})}{\cos 30^{\circ} \cos 40^{\circ} \cos 50^{\circ} \cos 60^{\circ}}$
= $\frac{\cos 60^{\circ} \cos 50^{\circ} \cos 40^{\circ} \cos 30^{\circ}}{\cos 30^{\circ} \cos 40^{\circ} \cos 50^{\circ} \cos 60^{\circ}}$
= $1$
Hence, the correct answer is $1$.