JEE Main 202126 Aug, Evening shiftMathematicsEasyTrigonometry118× since 2002If ∑r=150tan−112r2=p\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}{1 \over {2{r^2}}} = p}r=1∑50tan−12r21=p, then the value of tan p is :AOption A. 101102{{101} \over {102}}102101BOption B. 5051{{50} \over {51}}5150COption C. 100DOption D. 5150{{51} \over {50}}5051CheckSkip · show solution
