Q29 / 75 · JEE Main 2025 (29 Jan Shift 1) · Mathematics

Define a relation R on the interval \(\left[0, \frac{\pi}{2}\right)\) by \(x \mathrm{R} y\) if and only if \(\sec ^2 x-\tan ^2 y=1\). Then R is :

  • A
    both reflexive and transitive but not symmetric
  • Correct answer
    B
    an equivalence relation
  • C
    reflexive but neither symmetric not transitive
  • D
    both reflexive and symmetric but not transitive

Solution

$\begin{aligned} & \sec ^2 x-\tan ^2 x=1 \quad(\text { on replacing } y \text { with } x) \\ & \Rightarrow \text { Reflexive } \\ & \sec ^2 x-\tan ^2 y=1 \\ & \Rightarrow 1+\tan ^2 x+1-\sec ^2 y=1 \\ & \Rightarrow \sec ^2 y-\tan ^2 x=1 \\ & \Rightarrow \text { symmetric } \\ & \sec ^2 x-\tan ^2 y=1 \\ & \sec ^2 y-\tan ^2 z=1 \end{aligned}$ Adding both $\begin{aligned} & \Rightarrow \sec ^2 x-\tan ^2 y+\sec ^2 y-\tan ^2 z=1+1 \\ & \sec ^2 x+1-\tan ^2 z=2 \\ & \sec ^2 x-\tan ^2 z=1 \\ & \Rightarrow \text { Transitive } \end{aligned}$ hence equivalence releation

Hence, the correct answer is option: 'B'.

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