Evaluate: sqrt{3} operatorname{cosec} 20^{cir | vedclass

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Question

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

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) 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 $2$: $= \frac{2 \left( \frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ} ight)}{\frac{1}{2} (2 \sin 20^{\circ} \cos 20^{\circ})}$ Using $\sin(A - B) = \sin A \cos B - \cos A \sin B$ and $\sin 2	heta = 2 \sin 	heta \cos 	heta$: $= \frac{2 (\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ})}{\frac{1}{2} \sin 40^{\circ}}$ $= \frac{2 \sin(60^{\circ} - 20^{\circ})}{\frac{1}{2} \sin 40^{\circ}} = \frac{2 \sin 40^{\circ}}{\frac{1}{2} \sin 40^{\circ}} = 4$

Evaluate: $\sqrt{3} \operatorname{cosec} 20^{\circ} - \sec 20^{\circ}$

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