sqrt{3} operatorname{cosec} 20^{circ} - sec 2 | vedclass

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Question

(D) We have the 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) We have the 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 the numerator and denominator by $2$: $= \frac{2(\frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ})}{\sin 20^{\circ} \cos 20^{\circ}}$ $= \frac{2(\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ})}{\frac{1}{2} \sin 40^{\circ}}$ $= \frac{4 \sin(60^{\circ} - 20^{\circ})}{\sin 40^{\circ}}$ $= \frac{4 \sin 40^{\circ}}{\sin 40^{\circ}} = 4$

$\sqrt{3} \operatorname{cosec} 20^{\circ} - \sec 20^{\circ}$ is equal to

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