Prove

Let $\cos ^{-1} \frac{4}{5}=x$. Then, $\cos x=\frac{4}{5} \Rightarrow \sin x=\sqrt{1-\left(\frac{4}{5}\right)^{2}}=\frac{3}{5}$.

$\therefore \tan x=\frac{3}{4} \Rightarrow x=\tan ^{-1} \frac{3}{4}$

$\therefore \cos ^{-1} \frac{4}{5}=\tan ^{-1} \frac{3}{4}$....(1)

Now, let $\cos ^{-1} \frac{12}{13}=y .$ Then, $\cos y=\frac{12}{13} \Rightarrow \sin y=\frac{5}{13}$.

$\therefore \tan y=\frac{5}{12} \Rightarrow y=\tan ^{-1} \frac{5}{12}$

$\therefore \cos ^{-1} \frac{12}{13}=\tan ^{-1} \frac{5}{12}$....(2)

Let $\cos ^{-1} \frac{33}{65}=z$. Then, $\cos z=\frac{33}{65} \Rightarrow \sin z=\frac{56}{65}$.

$\therefore \tan z=\frac{56}{33} \Rightarrow z=\tan ^{-1} \frac{56}{33}$

$\therefore \cos ^{-1} \frac{33}{65}=\tan ^{-1} \frac{56}{33}$....(3)

Now, we will prove that:

L.H.S. $=\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}$

$=\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{5}{12}$      $[$ Using $(1)$ and $(2)]$

$=\tan ^{-1} \frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \cdot \frac{5}{12}}$    $\left[\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}\right]$

$=\tan ^{-1} \frac{36+20}{48-15}$

$=\tan ^{-1} \frac{56}{33}$

$=\tan ^{-1} \frac{56}{33}$   [by (3)]

$=$ R.H.S.