Solution:

Put that $Cos^{-1}\big(\frac{4}{5}\big) = A$

$Cos A = \frac{4}{5}$

$Sin A = \sqrt{1 - cos^{2}A}$

$ = \sqrt{1 - \frac{4}{5}}$

$ = \frac{4}{5}$

$Put Cos^{-1}\big(\frac{12}{13}\big) = B$

$Cos B = \frac{12}{13}$

$Sin B = \sqrt{1 - cos^{2}B}$

$ = \sqrt{1 - \frac{144}{169}}$

$ Sin B = \frac{5}{13}$

Consider,

Cos(A + B) = cosA.cosB - sinA.sinB

$Cos(A + B) = \big(\frac{4}{5}\big) \big(\frac{12}{13}\big) - \big(\frac{3}{5}\big) \big(\frac{5}{13}\big)$

$Cos(A + B) = \frac{33}{65}$

$A + B = Cos^{-1}\big(\frac{33}{65}\big)$

$Cos^{-1}\big(\frac{4}{5}\big) + Cos^{-1}\big(\frac{12}{13}\big) = Cos^{-1}\big(\frac{33}{65}\big)$