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(D) Given relation is defined as $	heta R \phi$ such that $\sec^{2} 	heta - 	an^{2} \phi = 1$.$1$. Reflexive: For any $	heta$, $	heta R 	heta$ i

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an equivalence relation

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(D) Given relation is defined as $	heta R \phi$ such that $\sec^{2} 	heta - 	an^{2} \phi = 1$.$1$. Reflexive: For any $	heta$, $	heta R 	heta$ implies $\sec^{2} 	heta - 	an^{2} 	heta = 1$. Since $1 + 	an^{2} 	heta = \sec^{2} 	heta$, this is $1 = 1$, which is true. Thus, $R$ is reflexive.$2$. Symmetric: If $	heta R \phi$, then $\sec^{2} 	heta - 	an^{2} \phi = 1$. Using $\sec^{2} 	heta = 1 + 	an^{2} 	heta$ and $	an^{2} \phi = \sec^{2} \phi - 1$, we get $(1 + 	an^{2} 	heta) - (\sec^{2} \phi - 1) = 1$, which simplifies to $	an^{2} 	heta - \sec^{2} \phi = -1$, or $\sec^{2} \phi - 	an^{2} 	heta = 1$. This implies $\phi R 	heta$. Thus, $R$ is symmetric.$3$. Transitive: If $	heta R \phi$ and $\phi R \psi$, then $\sec^{2} 	heta - 	an^{2} \phi = 1$ and $\sec^{2} \phi - 	an^{2} \psi = 1$. Adding these, $\sec^{2} 	heta - 	an^{2} \phi + \sec^{2} \phi - 	an^{2} \psi = 2$. Since $\sec^{2} \phi - 	an^{2} \phi = 1$, we have $\sec^{2} 	heta - 	an^{2} \psi + 1 = 2$, which means $\sec^{2} 	heta - 	an^{2} \psi = 1$. This implies $	heta R \psi$. Thus, $R$ is transitive.Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation.

For any two real numbers $\theta$ and $\phi$, we define $\theta R \phi$ if and only if $\sec^{2} \theta - \tan^{2} \phi = 1$. The relation $R$ is

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