MCQ
The value of $\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{p = 1}^n {2p} } \right)} } \right)$ is
  • $\frac{21}{19}$
  • B
    $\frac{19}{21}$
  • C
    $\frac{22}{23}$
  • D
    $\frac{23}{22}$

Answer

Correct option: A.
$\frac{21}{19}$
a
$\cot \left[ {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{p = 1}^n {2q} } \right)} } \right]$

$ = \cot \left[ {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + {n^2} + n} \right)} } \right]$

$ = \cot \left[ {\sum\limits_{n = 1}^{19} {{{\tan }^{ - 1}}\left( {\frac{1}{{1 + {n^2} + n}}} \right)} } \right]$

$ = \cot \left[ {\sum\limits_{n = 1}^{19} {{{\tan }^{ - 1}}\left( {n + 1} \right) - {{\tan }^{ - 1}}1} } \right]$

$ = \cot \left[ {{{\tan }^{ - 1}}20 - {{\tan }^{ - 1}}1} \right]$

$ = \cot \left( {{{\tan }^{ - 1}}\frac{{19}}{{21}}} \right)$

$ \Rightarrow \frac{{21}}{{19}}$

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