The value of frac{sqrt{3}text{cosec } 20^{cir | vedclass

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(C) Let $E = \frac{\sqrt{3}	ext{cosec } 20^{\circ}-\sec 20^{\circ}}{\cos 20^{\circ}\cos 40^{\circ}\cos 60^{\circ}\cos 80^{\circ}}$.First,simplify the

Vedclass · Accepted Answer

$64$

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(C) Let $E = \frac{\sqrt{3}	ext{cosec } 20^{\circ}-\sec 20^{\circ}}{\cos 20^{\circ}\cos 40^{\circ}\cos 60^{\circ}\cos 80^{\circ}}$.First,simplify the numerator: $\sqrt{3}	ext{cosec } 20^{\circ}-\sec 20^{\circ} = \frac{\sqrt{3}}{\sin 20^{\circ}} - \frac{1}{\cos 20^{\circ}} = \frac{\sqrt{3}\cos 20^{\circ} - \sin 20^{\circ}}{\sin 20^{\circ}\cos 20^{\circ}} = \frac{2(\frac{\sqrt{3}}{2}\cos 20^{\circ} - \frac{1}{2}\sin 20^{\circ})}{\sin 20^{\circ}\cos 20^{\circ}} = \frac{2\sin(60^{\circ}-20^{\circ})}{\sin 20^{\circ}\cos 20^{\circ}} = \frac{2\sin 40^{\circ}}{\sin 20^{\circ}\cos 20^{\circ}}$.Next,simplify the denominator: $\cos 20^{\circ}\cos 40^{\circ}\cos 60^{\circ}\cos 80^{\circ} = \cos 20^{\circ}\cos 40^{\circ} \cdot \frac{1}{2} \cdot \cos 80^{\circ} = \frac{1}{2} \cdot \frac{\sin(2^3 \cdot 20^{\circ})}{2^3 \sin 20^{\circ}} = \frac{1}{16} \cdot \frac{\sin 160^{\circ}}{\sin 20^{\circ}} = \frac{1}{16} \cdot \frac{\sin 20^{\circ}}{\sin 20^{\circ}} = \frac{1}{16}$.Thus,$E = \frac{2\sin 40^{\circ} / (\sin 20^{\circ}\cos 20^{\circ})}{1/16} = \frac{4\sin 40^{\circ} / \sin 40^{\circ}}{1/16} = 4 	imes 16 = 64$.

The value of $\frac{\sqrt{3}\text{cosec } 20^{\circ}-\sec 20^{\circ}}{\cos 20^{\circ}\cos 40^{\circ}\cos 60^{\circ}\cos 80^{\circ}}$ is equal to

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