Calculating $\cos^{-1}{\frac{3}{\sqrt10}} + \cos^{-1}{\frac{2}{\sqrt5}}$ $$\cos^{-1}{\frac{3}{\sqrt{10}}} + \cos^{-1}{\frac{2}{\sqrt 5}}= ?$$
Let $\cos^{-1}{\frac{3}{\sqrt{10}}}=\alpha,
 \cos^{-1}{\frac{2}{\sqrt 5}}=\beta$ then, $\cos\alpha=\frac{3}{\sqrt{10}}, \cos\beta=\frac{2}{\sqrt5}$
Therefore $$\cos\alpha=\frac{3\cdot2}{2\sqrt2\sqrt5}= \frac{3}{2\sqrt2}\cdot\cos\beta$$
This is all I did till now. Could you go further with this to answer?

Like Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $,
using $\cos(A+B)$ and the definition of principal values
$$\cos^{-1}x+\cos^{-1}y=\begin{cases}\cos^{-1}(xy-\sqrt{(1-x^2)(1-y^2)}) &\mbox{if }\cos^{-1}x+\cos^{-1}y\le\pi \\
2\pi-\cos^{-1}(xy-\sqrt{(1-x^2)(1-y^2)})  & \mbox{otherwise }  \end{cases}$$
Now $\cos^{-1}x+\cos^{-1}y\le\pi$ will happen
$\iff\cos^{-1}x\le\pi-\cos^{-1}y=\cos^{-1}(-y)$ using How do I prove that $\arccos(x) + \arccos(-x)=\pi$ when $x \in [-1,1]$?
$\iff\dfrac\pi2-\sin^{-1}x\le\dfrac\pi2-\sin^{-1}(-y)$
$\iff\sin^{-1}x\ge\sin^{-1}(-y)$
$\iff x\ge-y$
