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

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

JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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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^2} + n + 2}}} \right)

Option A (

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

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

n

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