Question

The value of $\frac{2\cos {40}^{\circ }-\cos {20}^{\circ }}{\sin {20}^{\circ }}=$

Answer 1

$\frac{2\cos {40}^{\circ }-\cos {20}^{\circ }}{\sin {20}^{\circ }}$

$\frac{\cos {40}^{\circ }+(-2\sin {30}^{\circ }.\sin {10}^{\circ })}{\sin {20}^{\circ }}=\frac{\cos {40}^{\circ }-\sin {10}^{\circ }}{\sin {20}^{\circ }}$

$\frac{\cos {40}^{\circ }-\cos {60}^{\circ }}{\sin {20}^{\circ }}$$=\frac{2\sin {60}^{\circ }.\sin {20}^{\circ }}{\sin {20}^{\circ }}$

$=2\times \frac{\sqrt{3}}{2}=\sqrt{3}$