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(D) The relation is defined as $P = \{(a,b) : \sec^2 a - 	an^2 b = 1\}$.$1$. Reflexive: For $P$ to be reflexive,$(a,a) \in P$ for all $a \in \mathbb{

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an equivalence relation

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(D) The relation is defined as $P = \{(a,b) : \sec^2 a - 	an^2 b = 1\}$.$1$. Reflexive: For $P$ to be reflexive,$(a,a) \in P$ for all $a \in \mathbb{R}$.Substituting $b=a$,we get $\sec^2 a - 	an^2 a = 1$,which is a standard trigonometric identity $1 + 	an^2 a - 	an^2 a = 1$. Thus,$1=1$ is true for all $a$. So,$P$ is reflexive.$2$. Symmetric: For $P$ to be symmetric,if $(a,b) \in P$,then $(b,a) \in P$.If $(a,b) \in P$,then $\sec^2 a - 	an^2 b = 1$. We check if $\sec^2 b - 	an^2 a = 1$.Using $\sec^2 x = 1 + 	an^2 x$,we have $\sec^2 b - 	an^2 a = (1 + 	an^2 b) - (\sec^2 a - 1) = 2 + 	an^2 b - \sec^2 a = 2 - (\sec^2 a - 	an^2 b) = 2 - 1 = 1$. Thus,$(b,a) \in P$. So,$P$ is symmetric.$3$. Transitive: For $P$ to be transitive,if $(a,b) \in P$ and $(b,c) \in P$,then $(a,c) \in P$.Given $\sec^2 a - 	an^2 b = 1$ and $\sec^2 b - 	an^2 c = 1$.We need to check if $\sec^2 a - 	an^2 c = 1$.From the first equation,$\sec^2 a = 1 + 	an^2 b$. From the second,$	an^2 c = \sec^2 b - 1$.Then $\sec^2 a - 	an^2 c = (1 + 	an^2 b) - (\sec^2 b - 1) = 2 + 	an^2 b - \sec^2 b = 2 - 1 = 1$. Thus,$(a,c) \in P$. So,$P$ is transitive.Since $P$ is reflexive,symmetric,and transitive,it is an equivalence relation.

Let $P$ be the relation defined on the set of all real numbers such that $P = \{(a,b) : \sec^2 a - \tan^2 b = 1\}$. Then $P$ is

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