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$\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}$ is equal to
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4
$\begin{aligned} & \sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}=4\left(\frac{\sqrt{3} \operatorname{cosec} 20^{\circ}}{4}-\frac{\sec 20^{\circ}}{4}\right) \\ & =4\left(\frac{\frac{\sqrt{3}}{2}}{2 \sin 20^{\circ}}-\frac{\frac{1}{2}}{2 \cos 20^{\circ}}\right)=4\left(\frac{\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ}}{2 \sin 20^{\circ} \cdot \cos 20^{\circ}}\right) \\ & =4\left(\frac{\sin 60^{\circ} \cdot \cos 20^{\circ}-\cos 60^{\circ} \cdot \sin 20^{\circ}}{\sin 40^{\circ}}\right) \\ & =4 \frac{\sin \left(60^{\circ}-20^{\circ}\right)}{\sin 40^{\circ}}=4 \frac{\sin 40^{\circ}}{\sin 40^{\circ}}=4\end{aligned}$
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