$P = \left\{ {\left( {a,b} \right):{{\sec }^2}a – {{\tan }^2}b = 1} \right\}$

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