The value of

$\frac{\cos 50^{\circ}}{\cos 130^{\circ}}=\frac{\cos 50^{\circ}}{\cos \left(180^{\circ}-50^{\circ}\right)}$

$=\frac{\cos 50^{\circ}}{-\cos 50^{\circ}} \quad\left(\because \cos \left(180^{\circ}-\theta\right)=-\cos \theta\right)$

$=-1$

Hence $\frac{\cos 50^{\circ}}{\cos 130^{\circ}}=-1$