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

both reflexive and transitive but not symmetric

both reflexive and symmetric but not transitive

reflexive but neither symmetric not transitive

Mathematics · Sets and Relations

JEE Main 2025 — 29 January, Morning Shift — Question 55

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

$\sec ^{2} x-\tan ^{2} x=1 \quad$ (on replacing $y$ with $x$ )

$\Rightarrow$ Reflexive $\sec ^{2} x-\tan ^{2} y=1$

$\Rightarrow 1+\tan ^{2} \mathrm{x}+1-\sec ^{2} \mathrm{y}=1$

$\Rightarrow \sec ^{2} y-\tan ^{2} \mathrm{x}=1$

$\Rightarrow$ symmetric

$\sec ^{2} x-\tan ^{2} y=1$ ,

$\sec ^{2} y-\tan ^{2} z=1$

Adding both $\Rightarrow \sec ^{2} x-\tan ^{2} y+\sec ^{2} y-\tan ^{2} z=1+1$

$\sec ^{2} x+1-\tan ^{2} z=2$

$\sec ^{2} x-\tan ^{2} z=1$

$\Rightarrow$ Transitive hence equivalence releation.

Answer key and solution verified before publishing.

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