If sin 10^{circ} sin 50^{circ} sin 60^{circ} | vedclass

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(B) First,calculate $m = \sin 10^{\circ} \sin 50^{\circ} \sin 60^{\circ} \sin 70^{\circ}$.Using the identity $\sin 	heta \sin(60^{\circ}-	heta) \sin

Vedclass · Accepted Answer

$16 \sqrt{3}$

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(B) First,calculate $m = \sin 10^{\circ} \sin 50^{\circ} \sin 60^{\circ} \sin 70^{\circ}$.Using the identity $\sin 	heta \sin(60^{\circ}-	heta) \sin(60^{\circ}+	heta) = \frac{1}{4} \sin 3	heta$,we have $\sin 10^{\circ} \sin 50^{\circ} \sin 70^{\circ} = \frac{1}{4} \sin(3 	imes 10^{\circ}) = \frac{1}{4} \sin 30^{\circ} = \frac{1}{4} 	imes \frac{1}{2} = \frac{1}{8}$.Since $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,then $m = \frac{1}{8} 	imes \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{16}$.Next,calculate $n = 	an 20^{\circ} 	an 40^{\circ} 	an 60^{\circ} 	an 80^{\circ}$.Using the identity $	an 	heta 	an(60^{\circ}-	heta) 	an(60^{\circ}+	heta) = 	an 3	heta$,we have $	an 20^{\circ} 	an 40^{\circ} 	an 80^{\circ} = 	an(3 	imes 20^{\circ}) = 	an 60^{\circ} = \sqrt{3}$.Since $	an 60^{\circ} = \sqrt{3}$,then $n = \sqrt{3} 	imes \sqrt{3} = 3$.Finally,$\frac{n}{m} = \frac{3}{\frac{\sqrt{3}}{16}} = \frac{3 	imes 16}{\sqrt{3}} = 16 \sqrt{3}$.

If $\sin 10^{\circ} \sin 50^{\circ} \sin 60^{\circ} \sin 70^{\circ} = m$ and $\tan 20^{\circ} \tan 40^{\circ} \tan 60^{\circ} \tan 80^{\circ} = n$,then find the value of $\frac{n}{m}$.

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