SOLUTION —
We have, $\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{14 \pi}{15}$
$\begin{array}{l}=\cos 24^{\circ} \cos 48^{\circ} \cos 96^{\circ} \cos 168^{\circ} \\=-\cos 12^{\circ} \cos 24^{\circ} \cos 48^{\circ} \cos 96^{\circ} \\=-\frac{16 \sin 12^{\circ}}{16 \sin 12^{\circ}} \cos 12^{\circ} \cos 24^{\circ} \cos 48^{\circ} \cos 96^{\circ} \\=\frac{-8 \sin 24^{\circ} \cos 24^{\circ} \cos 48^{\circ} \cos 96^{\circ}}{16 \sin 12^{\circ}} \\=\frac{-4 \sin 48^{\circ} \cos 48^{\circ} \cos 96^{\circ}}{16 \sin 12^{\circ}} \\=-\frac{2 \sin 96^{\circ} \cos 96^{\circ}}{16 \sin 12^{\circ}}=-\frac{\sin 192^{\circ}}{16 \sin 12^{\circ}} \\=\frac{\sin 12^{\circ}}{16 \sin 12^{\circ}}=\frac{1}{16} \\\end{array}$
So, The correct option will be (A).