Question

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

• A. $\frac{\sqrt{3}}{4}$
• B. $\frac{4}{\sqrt{3}}$
• C. $\frac{2}{\sqrt{3}}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

Answer: (B)

$\frac{1}{\cos 290^{\circ }}+\frac{1}{\sqrt{3}\sin 250^{\circ }}$
$=\frac{1}{\cos 70^{\circ }}-\frac{1}{\sqrt{3}\sin 110^{\circ }}$
$=\frac{\sqrt{3}\sin 110^{\circ }-\cos 70^{\circ }}{\sqrt{3}\sin 110^{\circ }\cos 70^{\circ }}$
$=\frac{\sqrt{3}\sin \left(180^{\circ }-70^{\circ }\right)-\cos 70^{\circ }}{\sqrt{3}\sin \left(180-70^{\circ }\right)\cos 70^{\circ }}$
$=\frac{\frac{\sqrt{3}}{2}\sin 70^{\circ }-\frac{1}{2}\cos 70^{\circ }}{\frac{\sqrt{3}}{2}\sin 70\cos 70^{\circ }}$
$=\frac{\cos 30^{\circ }\sin 70^{\circ }-\sin 30^{\circ }\cos 70^{\circ }}{\frac{\sqrt{3}}{2}\cdot \frac{1}{2}\sin 140^{\circ }}$
$=\frac{\sin \left(70^{\circ }-30^{\circ }\right)}{\frac{\sqrt{3}}{4}\sin \left(180^{\circ }-40^{\circ }\right)}$
$=\frac{\sin 40^{\circ }}{\frac{\sqrt{3}}{4}\sin 40^{\circ }}=\frac{4}{\sqrt{3}}$