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

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Question

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

Vedclass · Accepted Answer

$4$

Vedclass · Answer

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

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

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