JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Analyzing +4 | -1

The sum of series

{\tan ^{ - 1}}\frac{1}{{1 + 1 + {1^2}}} + {\tan ^{ - 1}}\frac{1}{{1 + 2 + {2^2}}} + {\tan ^{ - 1}}\frac{1}{{1 + 3 + {3^2}}} + \ldots

\infty

is equal to

View answer and explanation

Correct option: A

\frac{\pi }{4}

Option A (

\frac{\pi }{4}

) is correct because the given series telescopes to

{\tan ^{ - 1}}(\infty ) - {\tan ^{ - 1}}(1)

, which simplifies to

\frac{\pi }{2} - \frac{\pi }{4} = \frac{\pi }{4}

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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)