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A) Reflexive and symmetric but not transitive.

B) Reflexive and transitive but not symmetric.

C) Symmetric and transitive but not reflexive.

D) An equivalence relation.

Answer
Verified

Relation:-In mathematics relation defines the relationship between two different sets of information.

\[{\sec ^2}\theta = 1 + {\tan ^2}\theta \]

As in one of the options, equivalence relation is mentioned then surely we have to check all the three relations(reflexive, symmetric and transitive). So we will check the given relations one after the other.

\[P = [(a,b):{\sec ^2}a - {\tan ^2}b = 1]\]

\[ \Rightarrow P \in (a,b)\] Which means relation belongs to (a, b) now we have to prove relation also holds true for (b, a) to be symmetric.

\[ \Rightarrow {\sec ^2}a - {\tan ^2}b = 1\]

\[ \Rightarrow {\sec ^2}a = 1 + {\tan ^2}b\]

Now using identity \[{\sec ^2}\theta = 1 + {\tan ^2}\theta \]

\[ \Rightarrow {\sec ^2}a = {\sec ^2}b\] Or \[{\sec ^2}b = {\sec ^2}a\]

Therefore \[P \in (b,a)\] holds true and relation P is symmetric.

Now for $(a,a)$ to be in P we can write given relation as \[{\sec ^2}a - {\tan ^2}a = 1\]

\[ \Rightarrow {\sec ^2}a = 1 + {\tan ^2}a\] It is an identity so $(a,a) \in P$. Hence relation P is reflexive.

Now for transitive,

$P:a \to b \Rightarrow {\sec ^2}a = {\sec ^2}b$

$P:b \to c \Rightarrow {\sec ^2}b = {\sec ^2}c$

From both we can conclude that $P:a \to c \Rightarrow {\sec ^2}a = {\sec ^2}c$

Hence P is a transitive relation.

Therefore given relation is transitive, symmetric and reflexive, hence is an equivalence relation so correct option is (D)

* Reflexive relation:-In this every element maps to itself. Reflexive relation is given by

$(a,a) \in R$

* Symmetric relation:-In this if a=b then b=a is also true. In other words relation R is symmetric only if $(b,a) \in R$ is true when $(a,b) \in R$

* Transitive relation: -In this if $(x,y) \in R$, $(y,z) \in R$ then $(x,z) \in R$ will also be true.