Explanation

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