JEE Main 2019Inverse Trigonometric FunctionsInverse Trigonometric FunctionsHardQuestionThe value of cot(∑n=119cot−1(1+∑p=1n2p))\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}} \left( {1 + \sum\limits_{p = 1}^n {2p} } \right)} \right)cot(n=1∑19cot−1(1+p=1∑n2p)) is :OptionsA2223{{22} \over {23}}2322B2322{{23} \over {22}}2223C2119{{21} \over {19}}1921D1921{{19} \over {21}}2119Check AnswerHide SolutionSolutioncot(∑n=119cot−1(1+n(n+1))\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + n\left( {n + 1} \right)} \right.} } \right)cot(n=1∑19cot−1(1+n(n+1)) cot(∑n=119cot−1(n2+n+1))=cot(∑n=119tan−111+n(n+1))\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {{n^2} + n + 1} \right)} } \right) = \cot \left( {\sum\limits_{n = 1}^{19} {{{\tan }^{ - 1}}{1 \over {1 + n\left( {n + 1} \right)}}} } \right)cot(n=1∑19cot−1(n2+n+1))=cot(n=1∑19tan−11+n(n+1)1) ∑n=119(tan−1(n+1)−tan−1n)\sum\limits_{n = 1}^{19} {\left( {{{\tan }^{ - 1}}\left( {n + 1} \right) - {{\tan }^{ - 1}}n} \right)} n=1∑19(tan−1(n+1)−tan−1n) cot(tan−120−tan−11)=cotAcotβ+1cotβ−cotA\cot \left( {{{\tan }^{ - 1}}20 - {{\tan }^{ - 1}}1} \right) = {{\cot A\cot \beta + 1} \over {\cot \beta - \cot A}}cot(tan−120−tan−11)=cotβ−cotAcotAcotβ+1 (Where tanA === 20, tanB === 1) 1(120)+11−120=2119{{1\left( {{1 \over {20}}} \right) + 1} \over {1 - {1 \over {20}}}} = {{21} \over {19}}1−2011(201)+1=1921