Define a relation R on the interval [0, frac{ | vedclass

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(A) Given the relation $xRy \iff \sec^2 x - 	an^2 y = 1$ on the interval $[0, \frac{\pi}{2})$.$1$. Reflexive: For any $x \in [0, \frac{\pi}{2})$,we c

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an equivalence relation

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(A) Given the relation $xRy \iff \sec^2 x - 	an^2 y = 1$ on the interval $[0, \frac{\pi}{2})$.$1$. Reflexive: For any $x \in [0, \frac{\pi}{2})$,we check if $xRx$ holds.$\sec^2 x - 	an^2 x = 1$,which is a standard trigonometric identity.Thus,$R$ is reflexive.$2$. Symmetric: If $xRy$,then $\sec^2 x - 	an^2 y = 1$.Using the identity $\sec^2 	heta = 1 + 	an^2 	heta$,we have $(1 + 	an^2 x) - 	an^2 y = 1 \implies 	an^2 x = 	an^2 y$.Then $\sec^2 y - 	an^2 x = (1 + 	an^2 y) - 	an^2 x = 1 + 	an^2 x - 	an^2 x = 1$.Thus,$yRx$ holds,so $R$ is symmetric.$3$. Transitive: If $xRy$ and $yRz$,then $\sec^2 x - 	an^2 y = 1$ and $\sec^2 y - 	an^2 z = 1$.From the first,$	an^2 x = 	an^2 y$. From the second,$\sec^2 y = 1 + 	an^2 z$.Substituting $\sec^2 y = 1 + 	an^2 y$ into the second equation: $1 + 	an^2 y - 	an^2 z = 1 \implies 	an^2 y = 	an^2 z$.Since $	an^2 x = 	an^2 y$ and $	an^2 y = 	an^2 z$,we have $	an^2 x = 	an^2 z$.Then $\sec^2 x - 	an^2 z = (1 + 	an^2 x) - 	an^2 z = 1 + 0 = 1$.Thus,$xRz$ holds,so $R$ is transitive.Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

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

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