• # question_answer $\sqrt{3}\,\text{cosec}\,{{20}^{o}}-\sec \,{{20}^{o}}=$ [IIT 1988] A) 2 B) $\frac{2\,\sin {{20}^{o}}}{\sin {{40}^{o}}}$ C) 4 D) $\frac{4\,\sin {{20}^{o}}}{\sin {{40}^{o}}}$

$\sqrt{3}\text{cosec}\,20{}^\circ -\sec 20{}^\circ =\frac{\sqrt{3}}{\sin 20{}^\circ }-\frac{1}{\cos \,20{}^\circ }$ $=\frac{\sqrt{3}\cos 20{}^\circ -\sin 20{}^\circ }{\sin 20{}^\circ \cos 20{}^\circ }=\frac{2\left[ \frac{\sqrt{3}}{2}\cos 20{}^\circ -\frac{1}{2}\sin \,20{}^\circ \right]}{\frac{2}{2}\sin 20{}^\circ \cos 20{}^\circ }$ $=\frac{4\cos (20{}^\circ +30{}^\circ )}{\sin 40{}^\circ }=\frac{4\cos 50{}^\circ }{\sin 40{}^\circ }=\frac{4\sin 40{}^\circ }{\sin 40{}^\circ }=4$.