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  • Let P be the relation defined on the set of all real numbers such that <img alt="P=\left \{ \left ( a,b \right ) :sec^{2}a-\tan ^{2}b=1\right \}" src="https://learn.careers360.com/latex-i

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

Option: 1

 reflexive and symmetric but not transitive.
 


Option: 2

reflexive and transitive but not symmetric.


Option: 3

  symmetric and transitive but not reflexive.

 


Option: 4

an equivalence relation.


Answers (1)

\\P=(a, b): \quad \sec ^{2} a-\tan ^{2} b=1 \\ \\\text {For reflexive: } \\ \\\quad (a, a): \quad \sec ^{2} a-\tan ^{2} a=1 \\ \\(b, b): \quad \sec ^{2} b-\tan ^{2} b=1

Hence P is Reflexive

\\ \text {For Symmetric: } \\ \\\quad P(a, b): \quad \sec ^{2} a-\tan ^{2} b=1\quad\quad...(1) \\\\ P (b, a): \quad \sec ^{2} b-\tan ^{2} a=1

Using Equation 1

\\\left(1+\tan ^{2} a\right)-\left(\sec ^{2} b-1\right)=1 \\ \\\tan ^{2} a-\sec ^{2} b+2=1 \\ \\ \sec ^{2} b-\tan ^{2} a=1 \\

Hence P is symmetric

\\\text {For Transitivity: } \\\\ \quad P: (a, b): \sec ^{2} a-\tan ^{2} b=1 \quad\quad...(1) \\ \\\quad P: (b,c): \ \sec ^{2} b-\tan ^{2} c=1 \quad\quad...(2)

Adding Eq 1 and 2

\\(\sec ^{2} a-\tan ^{2} b)+(\sec ^{2} b-\tan ^{2} c)=2 \\

\\(\sec ^{2} a-\tan ^{2} c)+(\sec ^{2} b-\tan ^{2} b)=2 \\

\\(\sec ^{2} a-\tan ^{2} c)+1=2 \\

\\\sec ^{2} a-\tan ^{2} c=1 \\

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

Hence, P is Transistive

 

So P is an equivalence relation

Posted by

Sumit Saini

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