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)

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