Let P be the relation defined on the set of all real numbers such...

Let $P$ be the relation defined on the set of all real numbers such that $\text{P} = \left\{\left(\text{a, b}\right) \text{:} \sec^{2} \text{a} - \tan^{2} \text{b} = 1\right\}$ . Then $P$ is

Options

Areflexive and symmetric but not transitive
Bsymmetric and transitive but not reflexive
Creflexive and transitive but not symmetric
Dan equivalence relation

Correct answer

D. an equivalence relation

Step-by-step solution

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

\sec^{2} \text{a} - \tan^{2} \text{b} = 1

⇒ \sec^{2} \text{a} = 1 + \tan^{2} \text{b}

⇒ \sec^{2} \text{a} = \sec^{2} \text{b}

⇒ \left|\text{sec a}\right| = \left|\text{sec b}\right|

\sec^{2} \text{a} - \tan^{2} \text{b} = 1

1 + \tan^{2} \text{a} - \sec^{2} \text{b} + 1 = 1

\sec^{2} \text{b} - \tan^{2} \text{a} = 1

Hence, it is Symmetric.

If \left|\text{sec a}\right| = \left|\text{sec b}\right|

and \left|\text{sec b}\right| = \left|\text{sec c}\right|

then \left|\text{sec a}\right| = \left|\text{sec c}\right| ⇒

transitive <br/><br/>So it is an equivalence relation.<br/>

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