JEE Main 2018Inverse Trigonometric FunctionsInverse Trigonometric FunctionsHardQuestionThe value of cot(∑n=150tan−1(11+n+n2))\cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{1 \over {1 + n + {n^2}}}} \right)} } \right)cot(n=1∑50tan−1(1+n+n21)) is :OptionsA2625{{26} \over {25}}2526B2526{{25} \over {26}}2625C5051{{50} \over {51}}5150D5251{{52} \over {51}}5152Check AnswerHide SolutionSolutioncot(∑n=150tan−1(11+n+n2))\cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{1 \over {1 + n + {n^2}}}} \right)} } \right)cot(n=1∑50tan−1(1+n+n21)) =cot(∑n=150tan−1((n+1)−n1+(n+1)n)) = \cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{{(n + 1) - n} \over {1 + (n + 1)n}}} \right)} } \right)=cot(n=1∑50tan−1(1+(n+1)n(n+1)−n)) =cot(∑n=150(tan−1(n+1)−tan−1n) = \cot \left( {\sum\limits_{n = 1}^{50} {({{\tan }^{ - 1}}(n + 1) - {{\tan }^{ - 1}}n} } \right)=cot(n=1∑50(tan−1(n+1)−tan−1n) =cot(tan−151−tan−11) = \cot ({\tan ^{ - 1}}51 - {\tan ^{ - 1}}1)=cot(tan−151−tan−11) =cot(tan−1(51−11+51)) = \cot \left( {{{\tan }^{ - 1}}\left( {{{51 - 1} \over {1 + 51}}} \right)} \right)=cot(tan−1(1+5151−1)) =cot(cot−1(5250)) = \cot \left( {{{\cot }^{ - 1}}\left( {{{52} \over {50}}} \right)} \right)=cot(cot−1(5052)) =2625 = {{26} \over {25}}=2526