$\mathop \sum \limits_{m = 1}^n {\tan ^{ - 1}}\left( {\frac{{2m}}{{{m^4} + {m^2} + 2}}} \right)\;$ 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 is correct because the given summation can be simplified using the arctangent subtraction formula and telescoping series, resulting in the expression for option A.

More Questions from INVERSE TRIGONOMETRY

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The value of

\cos ^{-1}(\cos 12)-\sin ^{-1}(\sin 12)

is- The value of

\sin \left(2 \tan ^{-1}(1 / 3)\right)+

\cos \left(\tan ^{-1} 2 \sqrt{2}\right)

is Find

\cos ^ {-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right\}=

\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1

, then

x

is equal to If

{x^2} + {y^2} + {z^2} = {r^2},

then

{\tan ^{ - 1}}\left( {\frac{{xy}}{{zr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{yz}}{{xr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{xz}}{{yr}}} \right)\;

is equal to

2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)

is equal to