The value of sqrt{3} operatorname{cosec} 20^{ | vedclass

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(B) Given expression is $\sqrt{3} \operatorname{cosec} 20^{\circ} - \sec 20^{\circ}$.$= \frac{\sqrt{3}}{\sin 20^{\circ}} - \frac{1}{\cos 20^{\circ}} =

Vedclass · Accepted Answer

$4$

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(B) Given expression is $\sqrt{3} \operatorname{cosec} 20^{\circ} - \sec 20^{\circ}$.$= \frac{\sqrt{3}}{\sin 20^{\circ}} - \frac{1}{\cos 20^{\circ}} = \frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$Multiply the numerator and denominator by $2$:$= \frac{2 [(\frac{\sqrt{3}}{2}) \cos 20^{\circ} - (\frac{1}{2}) \sin 20^{\circ}]}{\sin 20^{\circ} \cos 20^{\circ}}$Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we have $\sin 20^{\circ} \cos 20^{\circ} = \frac{1}{2} \sin 40^{\circ}$.Also,$\frac{\sqrt{3}}{2} = \sin 60^{\circ}$ and $\frac{1}{2} = \cos 60^{\circ}$.$= \frac{2 [\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ}]}{\frac{1}{2} \sin 40^{\circ}}$Using the formula $\sin(A - B) = \sin A \cos B - \cos A \sin B$:$= \frac{2 \sin(60^{\circ} - 20^{\circ})}{\frac{1}{2} \sin 40^{\circ}} = \frac{2 \sin 40^{\circ}}{\frac{1}{2} \sin 40^{\circ}} = 4$.

The value of $\sqrt{3} \operatorname{cosec} 20^{\circ} - \sec 20^{\circ} = $

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