A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{3}\]
C) \[\frac{\pi }{2}\]
D) \[\frac{\pi }{5}\]
Correct Answer: A
Solution :
\[{{\cot }^{-1}}({{2.1}^{2}})+co{{t}^{-1}}({{2.2}^{2}})+{{\cot }^{-1}}({{2.3}^{2}})+....\infty \] \[\underset{r=1}{\mathop{\overset{\infty }{\mathop{\Sigma }}\,}}\,{{\cot }^{-1}}(2.\,{{r}^{2}})=\underset{r=1}{\mathop{\overset{\infty }{\mathop{\Sigma }}\,}}\,\,{{\tan }^{-1}}\left( \frac{1}{2{{r}^{2}}} \right)\] \[=\underset{r=1}{\mathop{\overset{\infty }{\mathop{\Sigma }}\,}}\,\,{{\tan }^{-1}}\left( \frac{(1+2r)+(1-2r)}{1-(1+2r)(1-2r)} \right)\] \[=\underset{r=1}{\mathop{\overset{\infty }{\mathop{\Sigma }}\,}}\,\,[{{\tan }^{-1}}(1+2r)+{{\tan }^{-1}}(1-2r)]\] \[=ta{{n}^{-1}}\,3-{{\tan }^{-1}}1+{{\tan }^{-1}}5+{{\tan }^{-1}}3\] \[+ta{{n}^{-1}}7-{{\tan }^{-1}}5+....+{{\tan }^{-1}}\infty \] \[=-\frac{\pi }{4}+\frac{\pi }{2}=\frac{\pi }{4}\]
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