$$\cos^{-1}{\frac{3}{\sqrt{10}}} + \cos^{-1}{\frac{2}{\sqrt 5}}= ?$$
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?
Use trig identity: $\sin^2\theta+\cos^2\theta=1$ $$\implies\sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-\left(\frac{3}{\sqrt{10}}\right)^2}=\frac{1}{\sqrt{10}}\quad \forall \quad 0\le\alpha\le \frac{\pi}{2}$$ $$\implies \sin\beta=\sqrt{1-\cos^2\beta}=\sqrt{1-\left(\frac{2}{\sqrt{5}}\right)^2}=\frac{1}{\sqrt{5}}\quad \forall \quad 0\le\beta\le \frac{\pi}{2}$$
Now, use trig identity $$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$$ $$\cos(\alpha+\beta)=\frac{3}{\sqrt {10}}\frac{2}{\sqrt 5}-\frac{1}{\sqrt {10}}\frac{1}{\sqrt 5} =\frac{1}{\sqrt2}$$
$$\implies \alpha+\beta=\cos^{-1}\frac{1}{\sqrt 2}=\frac{\pi}{4} \quad \quad\left(\because \cos(\alpha+\beta)\in[-1,1]\right)$$ $$\therefore \cos^{-1}{\frac{3}{\sqrt {10}}} + \cos^{-1}{\frac{2}{\sqrt 5}}= \color{blue}{\frac{\pi}{4}}$$ Or alternatively use trig identity $$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$$ $$\sin(\alpha+\beta)=\frac{1}{\sqrt {10}}\frac{2}{\sqrt 5}+\frac{3}{\sqrt {10}}\frac{1}{\sqrt 5} =\frac{1}{\sqrt2}$$
$$\implies \alpha+\beta=\sin^{-1}\frac{1}{\sqrt 2}=\frac{\pi}{4} \quad \quad\left(\because \sin(\alpha+\beta)\in[-1,1]\right)$$ $$\therefore \cos^{-1}{\frac{3}{\sqrt {10}}} + \cos^{-1}{\frac{2}{\sqrt 5}}=\color{red}{\frac{\pi}{4}}$$
Hint : apply this formula:
$$\cos^{-1} x +\cos^{-1}y=\cos^{-1}[xy-\sqrt{(1-x^2)(1-y^2)}]$$
Put $x=\frac2{\sqrt {10}}$ and $y=\frac 2{\sqrt 5}$
Hint: Using your notation, try using trigonometric identities to expand $$\begin{align} \cos (\alpha+\beta) \\ \sin(\alpha+\beta) \end{align} $$
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 $x \in [-1,1]$?
$\iff\dfrac\pi2-\sin^{-1}x\le\dfrac\pi2-\sin^{-1}(-y)$
$\iff\sin^{-1}x\ge\sin^{-1}(-y)$
$\iff x\ge-y$
