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(D) The relation is defined as $R = \{(a, b) \mid \sin ^{2} a + \cos ^{2} b = 1\}$.$1.$ Reflexivity: For any $a \in \mathbb{R}$, we have $\sin ^{2} a

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an equivalence relation

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(D) The relation is defined as $R = \{(a, b) \mid \sin ^{2} a + \cos ^{2} b = 1\}$.$1.$ Reflexivity: For any $a \in \mathbb{R}$, we have $\sin ^{2} a + \cos ^{2} a = 1$. Thus, $(a, a) \in R$. So, $R$ is reflexive.$2.$ Symmetry: Let $(a, b) \in R$, then $\sin ^{2} a + \cos ^{2} b = 1$. Using $\sin ^{2} x = 1 - \cos ^{2} x$ and $\cos ^{2} x = 1 - \sin ^{2} x$, we get $(1 - \cos ^{2} a) + (1 - \sin ^{2} b) = 1$, which simplifies to $\sin ^{2} b + \cos ^{2} a = 1$. Thus, $(b, a) \in R$. So, $R$ is symmetric.$3.$ Transitivity: Let $(a, b) \in R$ and $(b, c) \in R$. Then $\sin ^{2} a + \cos ^{2} b = 1$ and $\sin ^{2} b + \cos ^{2} c = 1$. Adding these equations: $\sin ^{2} a + \cos ^{2} b + \sin ^{2} b + \cos ^{2} c = 1 + 1 = 2$. Since $\sin ^{2} b + \cos ^{2} b = 1$, we have $\sin ^{2} a + 1 + \cos ^{2} c = 2$, which implies $\sin ^{2} a + \cos ^{2} c = 1$. Thus, $(a, c) \in R$. So, $R$ is transitive.Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation.

For any two real numbers $a$ and $b$, we define $a R b$ if and only if $\sin ^{2} a+\cos ^{2} b=1$. The relation $R$ is

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