Verify the equality without using tables: $\;\;\;$ $\dfrac{1}{\cos 290^\circ} + \dfrac{1}{\sqrt{3} \sin 250^\circ} = \dfrac{4}{\sqrt{3}}$
$\cos 290^\circ = \cos \left(270^\circ + 20^\circ\right) = \sin 20^\circ$
$\sin 250^\circ = \sin \left(270^\circ - 20^\circ\right) = - \cos 20^\circ$
$\begin{aligned}
LHS & = \dfrac{1}{\cos 290^\circ} + \dfrac{1}{\sqrt{3} \sin 250^\circ} \\\\
& = \dfrac{1}{\sin 20^\circ} - \dfrac{1}{\sqrt{3} \cos 20^\circ} \\\\
& = \dfrac{\sqrt{3} \cos 20^\circ - \sin 20^\circ}{\sqrt{3} \sin 20^\circ \cos 20^\circ} \\\\
& = \dfrac{2 \left(\dfrac{\sqrt{3}}{2} \cos 20^\circ - \dfrac{1}{2} \sin 20^\circ\right)}{\dfrac{\sqrt{3}}{2} \times 2 \sin 20^\circ \cos 20^\circ} \\\\
& = \dfrac{4 \left(\sin 60^\circ \cos 20^\circ - \cos 60^\circ \sin 20^\circ\right)}{\sqrt{3} \sin 40^\circ} \\\\
& = \dfrac{4 \sin \left(60^\circ - 20^\circ\right)}{\sqrt{3} \sin 40^\circ} \\\\
& = \dfrac{4 \sin 40^\circ}{\sqrt{3} \sin 40^\circ} \\\\
& = \dfrac{4}{\sqrt{3}} = RHS
\end{aligned}$
Hence verified.