STD 12 - 2. inverse trigonometric function — JEE कक्षा 11 साइन्स — Question

$ = \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}}$