Let $\cos^{-1}\frac{4}{5} = x$ and $\cos^{-1}\frac{12}{13} = y$. Then $\cos x = \frac{4}{5}$ and $\cos y = \frac{12}{13}$.

We find the sine values using the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$. Since $x, y \in [0, \pi]$, sine values are non-negative. Thus,

$$\sin x = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$$

and

$$\sin y = \sqrt{1 - \frac{144}{169}} = \frac{5}{13}$$

Consider the cosine of the sum using the cosine addition formula:

$$\cos(x+y) = \cos x \cos y - \sin x \sin y$$

Substituting the values, we get

$$\cos(x+y) = \left(\frac{4}{5}\right)\left(\frac{12}{13}\right) - \left(\frac{3}{5}\right)\left(\frac{5}{13}\right)$$

Simplifying the expression,

$$\cos(x+y) = \frac{48}{65} - \frac{15}{65} = \frac{33}{65}$$

Therefore, $x+y = \cos^{-1}\frac{33}{65}$.

Since $x$ and $y$ are in the principal branch $[0, \pi]$, and $\cos x, \cos y > 0$, both $x$ and $y$ lie in $(0, \pi/2)$. Thus, $x+y \in (0, \pi)$, which lies within the range of the principal branch of $\cos^{-1}$.

Substituting back $x$ and $y$, we get

$$\cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} = \cos^{-1}\frac{33}{65}$$

∴ Hence proved.