A) 2
B) \[\frac{2\,\sin {{20}^{o}}}{\sin {{40}^{o}}}\]
C) 4
D) \[\frac{4\,\sin {{20}^{o}}}{\sin {{40}^{o}}}\]
Correct Answer: C
Solution :
\[\sqrt{3}\text{cosec}\,20{}^\circ -\sec 20{}^\circ =\frac{\sqrt{3}}{\sin 20{}^\circ }-\frac{1}{\cos \,20{}^\circ }\] \[=\frac{\sqrt{3}\cos 20{}^\circ -\sin 20{}^\circ }{\sin 20{}^\circ \cos 20{}^\circ }=\frac{2\left[ \frac{\sqrt{3}}{2}\cos 20{}^\circ -\frac{1}{2}\sin \,20{}^\circ \right]}{\frac{2}{2}\sin 20{}^\circ \cos 20{}^\circ }\] \[=\frac{4\cos (20{}^\circ +30{}^\circ )}{\sin 40{}^\circ }=\frac{4\cos 50{}^\circ }{\sin 40{}^\circ }=\frac{4\sin 40{}^\circ }{\sin 40{}^\circ }=4\].
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