JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

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Application +4 | -1

Sum of series

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

upto infinite terms, is :

View answer and explanation

Correct option: C

\frac{3 \pi}{4}

The correct answer is C (

\frac{3 \pi}{4}

) because the series can be written as a telescoping series using the arctangent subtraction formula, and the sum converges to

\frac{3\pi}{4}

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x

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

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