The value of the expression sqrt{3} operatorn | vedclass

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Question

(C) Given expression: $\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}$ $= \frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}}$ $= \fr

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given expression: $\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}$ $= \frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}}$ $= \frac{\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$ Multiply 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 $\sin(A-B) = \sin A \cos B - \cos A \sin B$ where $A=60^{\circ}$ and $B=20^{\circ}$: $= \frac{2(\sin 60^{\circ} \cos 20^{\circ}-\cos 60^{\circ} \sin 20^{\circ})}{\frac{1}{2}(2 \sin 20^{\circ} \cos 20^{\circ})}$ $= \frac{2 \sin(60^{\circ}-20^{\circ})}{\frac{1}{2} \sin 40^{\circ}} = \frac{4 \sin 40^{\circ}}{\sin 40^{\circ}} = 4$

The value of the expression $\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}$ is equal to

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