Question

$\frac{1}{\cos {290}^{\circ }}+\frac{1}{\sqrt{3}\sin {250}^{\circ }}$ equals:

• A. $\frac{2\sqrt{3}}{3}$
• B. $\frac{4}{\sqrt{3}}$
• C. $\sqrt{3}$
• D. none

Answer 1

The correct option is B $\frac{4}{\sqrt{3}}$

$\frac{\frac{\sqrt{3}}{2}\sin {250}^{\circ }+\frac{1}{2}\cos {290}^{\circ }}{\frac{\sqrt{3}}{4}\cos {290}^{\circ }+\sin {250}^{\circ }}$

$\sin (A+B)$

$\Rightarrow \sin A\cos B+\cos A\sin B$

$\Rightarrow \sin (A+B)-\sin (A-B)$

$\sin {540}^{\circ }-\sin {40}^{\circ }$

$\Rightarrow \frac{\frac{\sqrt{3}}{2}\sin {250}^{\circ }+\frac{1}{2}\cos {290}^{\circ }}{\frac{-\sqrt{3}}{4}\left(\sin {40}^{\circ }\right)}$

$\Rightarrow \frac{\sqrt{3}}{2}-\sin {70}^{\circ }+\frac{1}{2}\cos {70}^{\circ }$

$\Rightarrow \frac{\frac{1}{2}\cos {70}^{\circ }-\frac{\sqrt{3}}{2}\sin {70}^{\circ }}{\frac{-\sqrt{3}}{4}\sin \left({40}^{\circ }\right)}$

$\Rightarrow \frac{\sin ({30}^{\circ }-{70}^{\circ })}{\frac{-\sqrt{3}}{4}\sin \left({40}^{\circ }\right)}$

$\Rightarrow \frac{-sin{40}^{\circ }}{\frac{-\sqrt{3}}{4}\sin {40}^{\circ }}$

$\Rightarrow \frac{4}{\sqrt{3}}$