• # question_answer The value of$\cot \left( \sum\limits_{n=1}^{19}{{{\cot }^{-1}}}\left( 1+\sum\limits_{p=1}^{n}{2p} \right) \right)$is: 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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