A) 0
B) \[\pi /4\]
C) \[\pi /2\]
D) \[\pi \]
Correct Answer: B
Solution :
\[{{\tan }^{-1}}\frac{1}{2}+{{\tan }^{-1}}\frac{1}{3}={{\tan }^{-1}}\left( \frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\times \frac{1}{3}} \right)\] \[={{\tan }^{-1}}\left( \frac{5}{5} \right)={{\tan }^{-1}}1=\frac{\pi }{4}\].
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