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Define a relation R on the interval [0, π 2 ) by x R y if and only if sec ^2 x-tan ^2 y=1. Then R is :

JEE Main 2025 — Mathematics Algebra

2025mcqeasy

Define a relation R on the interval [0,π2)\left[0, \frac{\pi}{2}\right)[0,2π​) by xRyx \mathrm{R} yxRy if and only if sec⁡2x−tan⁡2y=1\sec ^2 x-\tan ^2 y=1sec2x−tan2y=1. Then R is :

Official previous-year question

Held on 29 Jan 2025 · Verified 6 Jul 2026.

Options

  1. A

    both reflexive and transitive but not symmetric

  2. B

    an equivalence relation

  3. C

    reflexive but neither symmetric not transitive

  4. D

    both reflexive and symmetric but not transitive

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