JEE Main 2019MathematicsInverse Trigonometric FunctionsMediumMCQ

JEE Main 2019 — Inverse Trigonometric Functions Question with Solution

JEE Main 2019 (10 Jan Shift 2)

Question

The value of cot⁡∑n=119cot-1⁡1+∑p=1n2p is:

Choose an option

Show full solutionCorrect option: A
Correct answer
A2119

Step-by-step explanation

We have, cot⁡∑n=119cot-1⁡1+∑p=1n2p

=cot∑n=119tan-111+nn+1

=cot∑n=119tan-1n+1-n1+nn+1

=cot⁡∑n=119(tan-1n+1-tan-1n)

=cot⁡tan-120-tan-11

=1tan(tan-120-tan-11)

=120-11+(20).1=2119

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About this question

This is a previous-year question from JEE Main 2019, covering the Inverse Trigonometric Functions chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.