JEE Main 2025 Jan 29 Shift 1, Mathematics Q11: Relations

Use the identity ${\sec }^2\theta -{\tan }^2\theta =1$ for all $\theta$ in the given interval.

Reflexive: $xRx⟺{\sec }^{2}x-{\tan }^{2}x=1$, which is true. $✓$

Symmetric: if ${\sec }^{2}x-{\tan }^{2}y=1$, then ${\tan }^{2}y={\sec }^{2}x-1={\tan }^{2}x$, hence ${\tan }^{2}x={\tan }^{2}y$. Then ${\sec }^{2}y-{\tan }^{2}x={\sec }^{2}y-{\tan }^{2}y=1$, so $yRx.$ $✓$

Transitive: $xRy$ gives ${\tan }^{2}x={\tan }^{2}y$, and $yRz$ gives ${\tan }^{2}y={\tan }^{2}z$; hence ${\tan }^{2}x={\tan }^{2}z$, and ${\sec }^{2}x-{\tan }^{2}z={\sec }^{2}x-{\tan }^{2}x=1.$ $✓$

All three properties hold, so $R$ is an equivalence relation.