If ${\tan }^{-1}(2)$ and ${\tan }^{-1}(3)$ are two angles of a triangle, then what is the third angle?

Let the three angles be $A,B$, and $C$.$A={\tan }^{-1}(2),B={\tan }^{-1}(3)$.Sum of angles in a triangle is $\pi$.
$$A+B+C=\pi ⟹C=\pi -(A+B)$$
Using the identity ${\tan }^{-1}x+{\tan }^{-1}y=\pi +{\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$ (when $xy>1$):
$$A+B=\pi +{\tan }^{-1}\left(\frac{2+3}{1-(2\times 3)}\right)$$
$$A+B=\pi +{\tan }^{-1}\left(\frac{5}{-5}\right)=\pi +{\tan }^{-1}(-1)$$
Since ${\tan }^{-1}(-1)=-\pi /4$:
$$A+B=\pi -\frac{\pi }{4}=\frac{3\pi }{4}$$
Now, $C=\pi -\frac{3\pi }{4}=\frac{\pi }{4}$.