Question

The value of the expression $\frac{1}{cos{290}^{\circ }}+\frac{1}{\sqrt{3}sin{250}^{\circ }}$ is equal to

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$\frac{1}{cos({270}^{\circ }+{20}^{\circ })}+\frac{1}{\sqrt{2}sin({270}^{\circ }-{20}^{\circ })}=\frac{1}{sin{20}^{\circ }}+\frac{1}{\sqrt{3}(-cos{20}^{\circ })}=\frac{\sqrt{3}cos{20}^{\circ }-sin{20}^{\circ }}{\sqrt{3}cos{20}^{\circ }cos{20}^{\circ }}\frac{\frac{\sqrt{3}}{2}cos{20}^{\circ }-\frac{1}{2}sin{20}^{\circ }}{\frac{\sqrt{3}}{4}sin{40}^{\circ }}=\frac{\frac{\sqrt{3}}{2}cos{20}^{\circ }-cos{60}^{\circ }sin{20}^{\circ }}{\frac{\sqrt{3}}{4}sin{40}^{\circ }}=\frac{sin{60}^{\circ }cos{20}^{\circ }-cos{60}^{\circ }sin{20}^{\circ }}{\frac{\sqrt{3}}{4}sin{40}^{\circ }}=\frac{sin{40}^{\circ }}{\frac{\sqrt{3}}{4}sin{40}^{\circ }}=\frac{4}{\sqrt{3}}$