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

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

My Attempt:
Let $\alpha=\cos^{-1}\frac{4}{5}\implies0<\alpha\leq\pi$ and $\beta=\cos^{-1}\frac{12}{13}\implies0<\beta<\pi$,
thus $0\leq\alpha+\beta\leq2\pi$
$$
\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\frac{4}{5}.\frac{12}{13}-\frac{3}{5}.\frac{5}{13}=\frac{33}{65}=\cos\Big[\cos^{-1}\frac{33}{65}\Big]\\
\implies \cos^{-1}\frac{33}{65}=2n\pi\pm(\alpha+\beta)
$$
case 1: If $0\leq\alpha+\beta\leq\pi$,
$$
\cos^{-1}\frac{33}{65}=\alpha+\beta=\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}
$$
case 2: If $\pi<\alpha+\beta\leq 2\pi$,
$$
\cos^{-1}\frac{33}{65}=2\pi-(\alpha+\beta)
$$
Is there any thing wrong with my approach and how do I eliminate case 2 in similar problems ?
 A: Because $$\cos\left(\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}\right)=\frac{4}{5}\cdot\frac{12}{13}-\frac{3}{5}\cdot\frac{5}{13}=\frac{33}{65}$$ and $$0^{\circ}<\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}<90^{\circ}.$$
A: Since $$\frac{\sqrt2}{2}<\frac{4}{5}<1\implies  \cos^{-1}\frac{\sqrt2}{2} >\cos^{-1}\frac{4}{5}> \cos^{-1} 1 $$
that is $$\frac{\pi}{4}>\cos^{-1}\frac{4}{5}> 0 $$
similarly  $$\frac{\pi}{4}>\cos^{-1}\frac{12}{13}> 0 $$
Hence, $$0<\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}<\frac{\pi}{2}$$
Using $$\cos(\cos^{-1}x) = x~~~and~~~~\sin(\cos^{-1}x) =\sqrt{1-\cos^2(\cos^{-1}x)} =\color{red}{\sqrt{1-x^2}}$$
you easily arrive at $$\cos\left(\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}\right)\\=\cos\left(\cos^{-1}\frac{12}{13}\right)\cos\left(\cos^{-1}\frac{4}{5}\right)-\sin\left(\cos^{-1}\frac{12}{13}\right)\sin\left(\cos^{-1}\frac{4}{5}\right)\\=\frac{4}{5}\cdot\frac{12}{13}-\frac{3}{5}\cdot\frac{5}{13}=\frac{33}{65}$$
Done
A: As $\cos(A+B)=?$
using Principal values,
$$0\le\arccos x,\arccos y,\arccos(xy-\sqrt{(1-x^2)(1-y^2)})\le\pi$$
$$\implies0\le\arccos x+\arccos y\le2\pi$$
$$\implies\arccos x+\arccos y$$
$$=\begin{cases}\arccos(xy-\sqrt{(1-x^2)(1-y^2)}) &\mbox{if } 0\le\arccos x+\arccos y\le\pi \\ 
2\pi-\arccos(xy-\sqrt{(1-x^2)(1-y^2)})  & \mbox{ otherwise }  \end{cases}$$
Now $0\le\arccos x+\arccos y\le\pi$
As $\arccos(u)$ lies in $\in[0,\pi],$
$\arccos x+\arccos y\ge0\forall  x,y\in[0,1]$
So, we only need $$\arccos x+\arccos y\le\pi\iff\arccos x\le\pi-\arccos y=\arccos(-y)$$
As $\arccos$ is decreasing function $\in[0,1],$ we need $$x\le-y$$
See also: Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $
A: When we multiply complex numbers, their angles are added. 
$(4+3i)$ and $(12+5i)$ have angles $\cos^{-1}(\frac{4}{5})$ and $\cos^{-1}(\frac{12}{13})$ respectively. 
$$(4+3i)(12+5i) = 33+56i$$ 
The argument of $33+56i$ is $\cos^{-1}(\frac{33}{65})$. 
