Question : What will be the value of $\frac{\sin 30^{\circ} \sin 40^{\circ} \sin 50^{\circ} \sin 60^{\circ}}{\cos 30^{\circ} \cos 40^{\circ} \cos 50^{\circ} \cos 60^{\circ}}$?
Option 1: $\frac{1}{\sqrt{2}}$
Option 2: $\sqrt{3}$
Option 3: $1$
Option 4: $\frac{1}{\sqrt{3}}$
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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 $1$.
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