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Let $P$ be the relation defined on the set of all real numbers such that
$P = \left\{ {\left( {a,b} \right):{{\sec }^2}\,a - {{\tan }^2}\,b = 1\,} \right\}$. Then $P$ is
reflexive and symmetric but not transitive
reflexive and transitive but not symmetric
symmetric and transitive but not reflexive
an equivalence relation
Solution
$P = \left\{ {\left( {a,b} \right):{{\sec }^2}a – {{\tan }^2}b = 1} \right\}$
For reflexive:
${\sec ^2}a – {\tan ^2}b = 1$
$\left( {true\,\,\forall \,\,a} \right)$
For symmetric:
${\sec ^2}b – {\tan ^2}a = 1$
$L.H.S$
$1 + {\tan ^2}b – \left( {{{\sec }^2}a – 1} \right)$
$ = 1 + {\tan ^2}b – {\sec ^2}a + 1$
$ = – \left( {{{\sec }^2}a – {{\tan }^2}b} \right) + 2$
$ = – 1 + 2 = 1$
So, Relation is symmetric
For transitive:
if ${\sec ^2}a – {\tan ^2}b = 1$ and ${\sec ^2}b – {\tan ^2}c = 1$
${\sec ^2}a – {\tan ^2}c = \left( {1 + {{\tan }^2}b} \right) – \left( {{{\sec }^2}b – 1} \right)$
$ = – {\sec ^2}b – {\tan ^2}b + 2$
$ = – 1 + 2 = 1$
So, Relation is transitive.
Hence, Relation $P$ is an equivalence relation
