A) \[\frac{21}{19}\]
B) \[\frac{19}{21}\]
C) \[\frac{22}{23}\]
D) \[\frac{23}{22}\]
Correct Answer: A
Solution :
\[\cot \left[ \sum\limits_{n=1}^{19}{{{\cot }^{-1}}}\left( 1+\sum\limits_{p=1}^{n}{2p} \right) \right]\] |
\[=\cot \left[ \sum\limits_{n=1}^{19}{{{\cot }^{-1}}}(1+{{n}^{2}}+n) \right]\] |
\[=\cot \left[ \sum\limits_{n=1}^{19}{ta{{n}^{-1}}}\left( \frac{1}{1+{{n}^{2}}+n} \right) \right]\] |
\[=\cot \left[ \sum\limits_{n=1}^{19}{ta{{n}^{-1}}}(n+1)-ta{{n}^{-1}}1 \right]\] |
\[=\cot [ta{{n}^{-1}}20-{{\tan }^{-1}}1]\] |
\[=\cot \left( {{\tan }^{-1}}\frac{19}{21} \right)\]\[\Rightarrow \] \[\frac{21}{19}.\] |
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