MathematicsEasy66×since 2002Q4123If ∑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 :A101102{{101} \over {102}}102101B5051{{50} \over {51}}5150C100D5150{{51} \over {50}}5051Check answerSkip