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

(A) Let $\cos ^{-1} \frac{4}{5} = x$. Then $\cos x = \frac{4}{5}$.
Since $\sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - (\frac{4}{5})^2} = \sqrt{\frac{9}{25}} = \frac{3}{5}$,we have $\tan x = \frac{\sin x}{\cos x} = \frac{3/5}{4/5} = \frac{3}{4}$.
Thus,$\cos ^{-1} \frac{4}{5} = \tan ^{-1} \frac{3}{4}$ $\dots (1)$.
Now,let $\cos ^{-1} \frac{12}{13} = y$. Then $\cos y = \frac{12}{13}$.
Since $\sin y = \sqrt{1 - (\frac{12}{13})^2} = \sqrt{\frac{25}{169}} = \frac{5}{13}$,we have $\tan y = \frac{5/13}{12/13} = \frac{5}{12}$.
Thus,$\cos ^{-1} \frac{12}{13} = \tan ^{-1} \frac{5}{12}$ $\dots (2)$.
Using the formula $\tan ^{-1} A + \tan ^{-1} B = \tan ^{-1} \left( \frac{A + B}{1 - AB} \right)$:
$L.H.S = \tan ^{-1} \frac{3}{4} + \tan ^{-1} \frac{5}{12} = \tan ^{-1} \left( \frac{\frac{3}{4} + \frac{5}{12}}{1 - \frac{3}{4} \times \frac{5}{12}} \right) = \tan ^{-1} \left( \frac{\frac{9+5}{12}}{1 - \frac{15}{48}} \right) = \tan ^{-1} \left( \frac{56/48}{33/48} \right) = \tan ^{-1} \frac{56}{33}$.
Converting $\tan ^{-1} \frac{56}{33}$ back to $\cos ^{-1}$ form,let $\theta = \tan ^{-1} \frac{56}{33}$,then $\tan \theta = \frac{56}{33}$.
Using the triangle method,the hypotenuse is $\sqrt{56^2 + 33^2} = \sqrt{3136 + 1089} = \sqrt{4225} = 65$.
Thus,$\cos \theta = \frac{33}{65}$,which implies $\theta = \cos ^{-1} \frac{33}{65} = R.H.S$.