Let P be the relation defined on the set of a | vedclass

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Question

$P = \left\{ {\left( {a,b} ight):{{\sec }^2}a - {{	an }^2}b = 1} ight\}$

For reflexive:

${\sec ^2}a - {	an ^2}b = 1$

$\left( {true\,\,\

Vedclass · Accepted Answer

an equivalence relation

Vedclass · Answer

$P = \left\{ {\left( {a,b} ight):{{\sec }^2}a - {{	an }^2}b = 1} ight\}$

For reflexive:

${\sec ^2}a - {	an ^2}b = 1$

$\left( {true\,\,\forall \,\,a} ight)$

For symmetric:

${\sec ^2}b - {	an ^2}a = 1$

$L.H.S$

$1 + {	an ^2}b - \left( {{{\sec }^2}a - 1} ight)$

$ = 1 + {	an ^2}b - {\sec ^2}a + 1$

$ =  - \left( {{{\sec }^2}a - {{	an }^2}b} ight) + 2$

$ =  - 1 + 2 = 1$

So, Relation is symmetric 

For transitive:

if ${\sec ^2}a - {	an ^2}b = 1$ and ${\sec ^2}b - {	an ^2}c = 1$

${\sec ^2}a - {	an ^2}c = \left( {1 + {{	an }^2}b} ight) - \left( {{{\sec }^2}b - 1} ight)$

$ =  - {\sec ^2}b - {	an ^2}b + 2$

$ =  - 1 + 2 = 1$

So, Relation is transitive.

Hence, Relation $P$ is an equivalence relation

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