Question

The value of $\sqrt{3}\text{cosec}{20}^{\circ }-\sec {20}^{\circ }$ is equal to

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

Answer 1

The correct option is B 4

$\sqrt{2}cosec{20}^0-sec{20}^0$
$\frac{\sqrt{3}}{sin{20}^0}-\frac{1}{cos20}$
$\frac{2(\frac{\sqrt{3}}{2}cos{20}^0-\frac{1}{2}sin{20}^0)}{sin20cos{20}^0}=\frac{2\times sin40}{2sin20cos20}\times 2$
$=4$