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

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Question

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

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 and divide by $2$: $= 2 \left( \frac{\frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}} ight)$ $= 2 \left( \frac{\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}} ight)$ $= 2 \left( \frac{\sin(60^{\circ} - 20^{\circ})}{\sin 20^{\circ} \cos 20^{\circ}} ight)$ $= 2 \left( \frac{\sin 40^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}} ight)$ $= 2 \left( \frac{\sin 40^{\circ}}{\frac{1}{2} \sin 40^{\circ}} ight)$ $= 2 	imes 2 = 4$

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

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