If ${\tan }^{-1}2, {\tan }^{-1}3$ are two angles of a triangle then third angle is

$$\begin{gathered} \text{ In a }\mathrm{△}\text{ Sum of three angles }={180}^{\circ } \\ \text{ Let third angle }={\tan }^{-1}\theta \\ {\tan }^{-1}3+{\tan }^{-1}2+{\tan }^{-1}\theta =\pi \\ {\tan }^{-1}a+{\tan }^{-1}b={\tan }^{-1}\frac{(a+b)}{(1-ab)} \\ \text{ So }{\tan }^{-1}\left(\frac{3+2}{1-3.2}\right)+{\tan }^{-1}\theta =\pi \\ {\tan }^{-1}(-1)+{\tan }^{-1}\theta =\pi \\ \text{ We know that }{\tan }^{-1}(-1)={135}^0\text{ or }\frac{3\pi }{4} \\ \Rightarrow \frac{3\pi }{4}+{\tan }^{-1}\left(\theta \right)=\pi \\ \Rightarrow {\tan }^{-1}\left(\theta \right)=\pi -\frac{3\pi }{4}=\frac{\pi }{4} \\ \text{ So, third angle }={\tan }^{-1}\left(\theta \right)={45}^{\circ }\text{ or }\frac{\pi }{4} \end{gathered}$$