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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Sets and Relations

NTA Abhyas 2020 · Mathematics · Sets and Relations

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

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

Options

  1. Areflexive and symmetric but not transitive
  2. Bsymmetric and transitive but not reflexive
  3. Creflexive and transitive but not symmetric
  4. 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\}<br/><br/><br /><br />\sec^{2} \text{a} - \tan^{2} \text{b} = 1<br/><br/><br /><br />⇒ \sec^{2} \text{a} = 1 + \tan^{2} \text{b}<br/><br/><br /><br />⇒ \sec^{2} \text{a} = \sec^{2} \text{b}<br/><br/><br /><br />⇒ \left|\text{sec a}\right| = \left|\text{sec b}\right|<br/><br/><br /><br />\sec^{2} \text{a} - \tan^{2} \text{b} = 1<br/><br/><br /><br />1 + \tan^{2} \text{a} - \sec^{2} \text{b} + 1 = 1<br/><br/><br /><br />\sec^{2} \text{b} - \tan^{2} \text{a} = 1<br/><br/>Hence,itisSymmetric.<br/><br/>If∣sec a∣=∣sec b∣<br /><br />Hence, it is Symmetric.<br /><br />If \left|\text{sec a}\right| = \left|\text{sec b}\right| and \left|\text{sec b}\right| = \left|\text{sec c}\right|then∣sec a∣=∣sec c∣⇒then \left|\text{sec a}\right| = \left|\text{sec c}\right| ⇒ transitive&

So it is an equivalence relation.
&
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