Question

For any two real number $a$and $b$, we defined $aRb$ if and only if ${\sin }^{2}a+{\cos }^{2}b=1$. The relation$R$ is

• A. reflexive but not symmetric
• B. symmetric but not transitive
• C. transitive but not reflexive
• D. an equivalence relation

Answer 1

The correct option is D

an equivalence relation

Explanation for the correct answer:

Option $\left(D\right)$: An equivalence relation

Given, $aRb\Rightarrow {\sin }^{2}a+{\cos }^{2}b=1$

Reflexive: A relation $R$ is said to be Reflexive when $R=\left\{\left(a,a\right)\in R,a\in A\right\}$

$aRa\Rightarrow {\sin }^{2}a+{\cos }^{2}a=1\forall a\in R$ which is True.

Therefore, relation $R$ is reflexive.

Symmetric: A relation $R$ is said to be Symmetric when $R=\left\{\left(a,b\right)\in R,\left(b,a\right)\in R\right\}$

$aRb\Rightarrow {\sin }^{2}a+{\cos }^{2}b=1$
$\Rightarrow 1–{\cos }^{2}a+1–{\sin }^{2}b=1(\because {\sin }^{2}a=1–{\cos }^{2}a\text{and}{\cos }^{2}b=1–{\sin }^{2}b)$
$\Rightarrow {\sin }^{2}b+{\cos }^{2}a=1$

Since, $bRa\forall a,b\in R$ is True

Therefore, relation $R$ is symmetric.

Transitive: A relation $R$ is said to be Transitive when $R=\left\{\left(a,b\right)\in R\left(b,c\right)\in R\left(a,c\right)\in R\right\}$

$aRa,bRc$
$\Rightarrow {\sin }^{2}a+{\cos }^{2}b=1$and ${\sin }^{2}b+{\cos }^{2}c=1$
Adding these two equations we get
$\Rightarrow {\sin }^{2}a+{\cos }^{2}b+{\sin }^{2}b+{\cos }^{2}c=2$
$\Rightarrow {\sin }^{2}a+{\cos }^{2}c=1(\because {\sin }^{2}b+{\cos }^{2}b=1)$

$\Rightarrow aRc$ is True
Since, $R$ being a reflexive, Symmetric and transitive relation.

Therefore, it is an equivalence relation.

Hence, option $\left(D\right)$ is the correct answer.