sqrt{3} csc 20^{circ} - sec 20^{circ} = | vedclass

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Question

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

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given expression: $\sqrt{3} \csc 20^{\circ} - \sec 20^{\circ} = \frac{\sqrt{3}}{\sin 20^{\circ}} - \frac{1}{\cos 20^{\circ}}$$= \frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$Multiply numerator and denominator by $2$:$= \frac{2 \left( \frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ} \right)}{\frac{1}{2} (2 \sin 20^{\circ} \cos 20^{\circ})}$Using $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$ and $\cos 60^{\circ} = \frac{1}{2}$:$= \frac{2 (\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ})}{\frac{1}{2} \sin 40^{\circ}}$$= \frac{4 \sin(60^{\circ} - 20^{\circ})}{\sin 40^{\circ}} = \frac{4 \sin 40^{\circ}}{\sin 40^{\circ}} = 4$.

$\sqrt{3} \csc 20^{\circ} - \sec 20^{\circ} = $

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