JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

1 of 1

Application +4 | -1

The value of

\cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {\frac{1}{{1 + n + {n^2}}}} \right)} } \right)

View answer and explanation

Correct option: A

\frac{{26}}{{25}}

Option A (

\frac{{26}}{{25}}

) is correct because the telescoping sum simplifies to

\cot \left( {{\tan }^{ - 1}}\left( {50} \right) - {{\tan }^{ - 1}}\left( 1 \right) \right)

, which evaluates to

\frac{{26}}{{25}}

More Questions from INVERSE TRIGONOMETRY

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

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

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

\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

x

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

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

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