JEE Main 2017MathematicsInverse Trigonometric FunctionsMediumMCQ

JEE Main 2017 — Inverse Trigonometric Functions Question with Solution

JEE Main 2017 (08 Apr Online)

Question

The value of tan-1⁡1+x2+ 1-x21+x2- 1-x2, x<12, x≠0, is equal to:

Choose an option

Show full solutionCorrect option: A
Correct answer
Aπ4+12cos-1⁡x2

Step-by-step explanation

Assume that ,
x2=cos⁡2θ;θ=12cos-1x2

∴tan- 1⁡1+cos⁡2θ+ 1-cos⁡2θ1+cos⁡2θ- 1-cos⁡2θ=tan- 1⁡2cos2θ+ 2sin2θ2cos2θ- 2sin2θ=tan- 1⁡cosθ+ sinθcosθ- sinθ
=tan-1⁡1+tan⁡θ1-tan⁡θ

=tan-1⁡tan⁡π4+θ

=π4+12cos-1⁡x2

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About this question

This is a previous-year question from JEE Main 2017, 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.