Q102 / 200 · WBJEE 2013 · Mathematics
For any two real numbers $a$ and $b$, we define $a R b$ if and only if $\sin ^{2} a+\cos ^{2} b=1$. The relation $R$ is
- Areflexive but not symmetric
- Bsymmetric but not transitive
- Ctransitive but not reflexive
- Correct answerDan equivalence relation
Solution
Let the given relation defined as
$R=\left\{(a, b) \mid \sin ^{2} a+\cos ^{2} b=1\right\}$
For reflexive, $\sin ^{2} a+\cos ^{2} a=1$
$\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1, \forall \theta \in R\right)$
$\Rightarrow { }_{a} R_{a} \Rightarrow(a, a) \in R$
$\Rightarrow R$ is reflexive.
For symmetric, $\sin ^{2} a+\cos ^{2} b=1$
$\Rightarrow 1-\cos ^{2} a+1-\sin ^{2} b=1$
$\Rightarrow \sin ^{2} b+\cos ^{2} a=1$
$\Rightarrow$ bRa
Hence, $R$ is symmetric.
For transitive
Let $a R b, b R c$ $\Rightarrow \sin ^{2} a+\cos ^{2} b=1$
and $\sin ^{2} b+\cos ^{2} c=1$
On adding Eqs. (0) and (ii), we get $\sin ^{2} a+\left(\sin ^{2} b+\cos ^{2} b\right)+\cos ^{2} c=2$
$\Rightarrow \sin ^{2} a+\cos ^{2} c+1=2$
$\Rightarrow \sin ^{2} a+\cos ^{2} c=1$
Hence, $A$ is transitive also. Therefore, relation $A$ is an equivalence relation.
$R=\left\{(a, b) \mid \sin ^{2} a+\cos ^{2} b=1\right\}$
For reflexive, $\sin ^{2} a+\cos ^{2} a=1$
$\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1, \forall \theta \in R\right)$
$\Rightarrow { }_{a} R_{a} \Rightarrow(a, a) \in R$
$\Rightarrow R$ is reflexive.
For symmetric, $\sin ^{2} a+\cos ^{2} b=1$
$\Rightarrow 1-\cos ^{2} a+1-\sin ^{2} b=1$
$\Rightarrow \sin ^{2} b+\cos ^{2} a=1$
$\Rightarrow$ bRa
Hence, $R$ is symmetric.
For transitive
Let $a R b, b R c$ $\Rightarrow \sin ^{2} a+\cos ^{2} b=1$
and $\sin ^{2} b+\cos ^{2} c=1$
On adding Eqs. (0) and (ii), we get $\sin ^{2} a+\left(\sin ^{2} b+\cos ^{2} b\right)+\cos ^{2} c=2$
$\Rightarrow \sin ^{2} a+\cos ^{2} c+1=2$
$\Rightarrow \sin ^{2} a+\cos ^{2} c=1$
Hence, $A$ is transitive also. Therefore, relation $A$ is an equivalence relation.
Hence, the correct answer is option: 'D'.