1. Let $O_1$ and $O_2$ be the centers of circles $\omega_1$ and $\omega_2$ respectively. Since each circle passes through the center of the other, the distance between $O_1$ and $O_2$ is equal to the radius $R$ of the circles. Thus, $O_1O_2 = R$.

2. Let $A, B, C$ be points on $\omega_1$ such that $AC$ and $BC$ are tangent to $\omega_2$ at points $D$ and $E$ respectively. Since $AC$ and $BC$ are tangents to $\omega_2$, we have $O_2D \perp AC$ and $O_2E \perp BC$.

3. Since $\omega_1$ and $\omega_2$ are equal circles and each passes through the center of the other, the triangles $O_1O_2A$ and $O_1O_2B$ are isosceles with $O_1A = O_1B = R$ and $O_2A = O_2B = R$.

4. Consider the angles $\angle O_2AC$ and $\angle O_2BC$. Since $O_2D \perp AC$ and $O_2E \perp BC$, the angles $\angle O_2AC$ and $\angle O_2BC$ are right angles. Therefore, $\angle O_2AC = \angle O_2BC = 90^\circ$.

5. Let $\theta = \angle O_1O_2A = \angle O_1O_2B$. Since $O_1O_2 = R$, the triangles $O_1O_2A$ and $O_1O_2B$ are congruent. Thus, $\angle O_1AO_2 = \angle O_1BO_2 = \theta$.

6. Since $\angle A + \angle B + \angle C = 180^\circ$ in triangle $ABC$, and $\angle A = \angle O_2AC$ and $\angle B = \angle O_2BC$, we have:
$$\angle A + \angle B = 180^\circ - \angle C$$

7. Using the cosine rule in triangle $O_1O_2A$, we have:
$$\cos \theta = \frac{O_1O_2^2 + O_1A^2 - O_2A^2}{2 \cdot O_1O_2 \cdot O_1A} = \frac{R^2 + R^2 - R^2}{2 \cdot R \cdot R} = \frac{R^2}{2R^2} = \frac{1}{2}$$

8. Since $\cos A = \cos \theta$ and $\cos B = \cos \theta$, we have:
$$\cos A + \cos B = \cos \theta + \cos \theta = 2 \cos \theta = 2 \cdot \frac{1}{2} = 1$$

The final answer is $\boxed{ \cos A + \cos B = 1 }$.