A) \[\frac{22}{23}\]
B) \[\frac{23}{22}\]
C) \[\frac{21}{19}\]
D) \[\frac{19}{21}\]
Correct Answer: C
Solution :
\[\sum\limits_{n\,=\,1}^{19}{{{\cot }^{-1}}\,\left( 1+\frac{2n(n+1)}{2} \right)}\] \[=\,\,{{\tan }^{-1}}\,\left( \frac{1}{1+n(n+1)} \right)\] \[=\,\,{{\tan }^{-1}}\,\left( \frac{(n+1)-n}{1+n(n+1)} \right)\] \[=\,\,\,\,\sum\limits_{n=1}^{19}{{{\tan }^{-1}}(n+1)\,-\,ta{{n}^{-1}}(n)}\] \[=ta{{n}^{-1}}(2)\,\,-ta{{n}^{-1}}(1)+ta{{n}^{-1}}(3)-ta{{n}^{-1}}(\,2)\] \[+....+\,\,ta{{n}^{-1}}(20)\,\,-ta{{n}^{-1}}(19)\] \[\cot \left[ \sum\limits_{n=1}^{19}{{{\cot }^{-1}}}\left( 1+\sum\limits_{p=1}^{n}{2p} \right) \right]\,\,=\,\,\cot (ta{{n}^{-1}}20-ta{{n}^{-1}}1)\]\[=\,\,\,\cot \left( {{\tan }^{-1}}\,\left( \frac{20-1}{1+20.1} \right) \right)\] \[=\,\,\,\cot \left( {{\tan }^{-1}}\,\left( \frac{19}{21} \right) \right)\] \[=\,\,\,\cot \left( {{\tan }^{-1}}\,\left( \frac{21}{19} \right) \right)\,\,=\,\,\frac{21}{19}\]
You need to login to perform this action.
You will be redirected in
3 sec
