Question

The value of the expression. $\sqrt{3}\text{cosec}{20}^0-\sec {20}^0$ is equal to

• A. $2$
• B. $\frac{2\sin {20}^0}{\sin {40}^0}$
• C. $4$
• D. $4\frac{\sin {20}^0}{\sin {40}^0}$

Answer 1

The correct option is C $4$

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

$=\frac{\sqrt{3}\cos {20}^{\circ }-\sin {20}^{\circ }}{\sin {20}^{\circ }\cos {20}^{\circ }}$

$=\frac{4(\frac{3}{\sqrt{2}}\cos {20}^{\circ }-\frac{1}{2}\sin {20}^{\circ })}{2\sin {20}^{\circ }\cos {20}^{\circ }}$

$=\frac{4\sin (60-20)}{\sin {40}^{\circ }}$

$=4$