JEE Main 202227 Jun, Evening shiftMathematicsMediumTrigonometry118× since 2002The 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 :AOption A. 2625{{26} \over {25}}2526BOption B. 2526{{25} \over {26}}2625COption C. 5051{{50} \over {51}}5150DOption D. 5251{{52} \over {51}}5152CheckSkip · show solution
