A) \[\frac{\pi }{3}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{6}\]
D) \[\frac{\pi }{2}\]
Correct Answer: B
Solution :
[b] We have \[A={{\tan }^{-1}}2\Rightarrow \tan A=2\] and \[B={{\tan }^{-1}}3\Rightarrow \tan B=3.\] Since, A, B, C are angles of a triangle \[\therefore A+B+C=\pi \] \[\Rightarrow C=\pi -(A+B)\] ? (1) Now, \[A+B={{\tan }^{-1}}2+{{\tan }^{-1}}3\] \[=\pi +{{\tan }^{-1}}\left( \frac{2+3}{1-2.3} \right)\] \[\left[ \because {{\tan }^{-1}}x+{{\tan }^{-1}}y=\pi +{{\tan }^{-1}}\frac{x+y}{1-xy} \right]\] \[=\pi +{{\tan }^{-1}}(-1)=\pi -ta{{n}^{-1}}(-1)\] \[=\pi -\frac{\pi }{4}=\frac{3\pi }{4}\] \[\therefore \] from (1), \[C=\pi -\frac{3\pi }{4}=\frac{\pi }{4}.\]
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