Question

What is $(\frac{sec{18}^{\circ }}{sec{144}^{\circ }}+\frac{cosec{18}^{\circ }}{cosec{144}^{\circ }})$ equals to?

• A. $\sec {18}^{\circ }$
• B. $\csc {18}^{\circ }$
• C. $-\sec {18}^{\circ }$
• D. $-\csc {18}^{\circ }$

Answer 1

The correct option is A $\sec {18}^{\circ }$

$\frac{\sec {18}^{\circ }}{\sec {144}^{\circ }}+\frac{\csc {18}^{\circ }}{\csc {144}^{\circ }}$

$=\frac{\cos {144}^{\circ }}{\cos {18}^{\circ }}+\frac{\sin {144}^{\circ }}{\sin {18}^{\circ }}$

$=\frac{\cos (180-{36}^{\circ })}{\cos {18}^{\circ }}+\frac{\sin (180-{36}^{\circ })}{\sin {18}^{\circ }}$

$=\frac{\sin {36}^{\circ }}{\sin {18}^{\circ }}-\frac{\cos {36}^{\circ }}{\cos {18}^{\circ }}$

$=2\cos {18}^{\circ }-\frac{{\cos }^2{18}^{\circ }-{\sin }^2{18}^{\circ }}{\cos {18}^{\circ }}$

$=\frac{{\cos }^2{18}^{\circ }+{\sin }^2{18}^{\circ }}{\cos {18}^{\circ }}$

$=\sec {18}^{\circ }$