$\mathop \sum \limits_{m = 1}^n {\tan ^{ - 1}}\left( {\frac{{2m}}{{{m^4} + {m^2} + 2}}} \right)$ is equal to

${\tan ^{ - 1}}\left( {\frac{{{n^2} - n}}{{{n^2} - n + 2}}} \right)$

${\tan ^{ - 1}}\left( {{n^2} + n + 2} \right)$

JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

1 of 1

Analyzing +4 | -1

\sum \limits_{m = 1}^n {\tan ^{ - 1}}\left( {\frac{{2m}}{{{m^4} + {m^2} + 2}}} \right)

is equal to

View answer and explanation

Correct option: A

{\tan ^{ - 1}}\left( {\frac{{{n^2} + n}}{{{n^4} + {n^2} + 2}}} \right)

Option D is incorrect. Option A (

{\tan ^{ - 1}}\left( {\frac{{{{{n^2} + n}}}}{{{{{n^4} + {n^2} + 2}}}}} \right)

) is the correct answer because the sum telescopes, leaving a simplified expression involving

n