In a triangle $$ABC,$$  if $$A = {\tan ^{ - 1}}2$$   and $$B = {\tan ^{ - 1}}3,$$   then $$C$$ is equal to

Solution :
$$\eqalign{ & {\text{We have, }}A = {\tan ^{ - 1}}2 \cr & \Rightarrow \tan A = 2\,{\text{and}}\,B = {\tan ^{ - 1}}3 \cr & \Rightarrow \tan B = 3. \cr & {\text{Since, }}A,B,C{\text{ are angles of a triangle}} \cr & \therefore A + B + C = \pi \cr & \Rightarrow C = \pi - \left( {A + B} \right)\,\,\,.....\left( 1 \right) \cr & {\text{Now}},A + B = {\tan ^{ - 1}}2 + {\tan ^{ - 1}}3 \cr & = \pi + {\tan ^{ - 1}}\left( {\frac{{2 + 3}}{{1 - 2 \cdot 3}}} \right)\,\left[ {\because {{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y = \pi + {{\tan }^{ - 1}}\frac{{x + y}}{{1 - xy}}} \right] \cr & = \pi + {\tan ^{ - 1}}\left( { - 1} \right) = \pi - {\tan ^{ - 1}}\left( { - 1} \right) \cr & = \pi - \frac{\pi }{4} = \frac{{3\pi }}{4} \cr & \therefore {\text{from}}\,\left( 1 \right),C = \pi - \frac{{3\pi }}{4} = \frac{\pi }{4}. \cr} $$