A) 1
B) 2
C) 3
D) 4
Correct Answer: B
Solution :
\[{{\tan }^{-1}}\left( \frac{4}{4{{r}^{2}}+3} \right)={{\tan }^{-1}}\left( \frac{1}{{{r}^{2}}+\frac{3}{4}} \right)\] \[={{\tan }^{-1}}\left( \frac{1}{1+\left( r+\frac{1}{2} \right)\left( r-\frac{1}{2} \right)} \right)\] \[={{\tan }^{-1}}\left( r+\frac{1}{2} \right)-{{\tan }^{-1}}\left( r-\frac{1}{2} \right)\] \[\therefore \sum\limits_{r=1}^{n}{{{\tan }^{-1}}}\left( \frac{4}{4{{r}^{2}}+3} \right)=\left( {{\tan }^{-1}}\left( n+\frac{1}{2} \right)-{{\tan }^{-1}}\frac{1}{2} \right)\] \[\Rightarrow \sum\limits_{r=1}^{\infty }{{{\tan }^{-1}}}\left( \frac{4}{4{{r}^{2}}+3} \right)={{\cot }^{-1}}\frac{1}{2}=({{\tan }^{-1}}2)\] \[\therefore \tan ({{\tan }^{-1}}2)=2\]
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