Question

For any two real numbers $a$ and $b$, we define $aRb$ if and only if $sin^2 a + cos^2b = 1$. 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

Answer: (D)

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. (i) and (ii), we get $\sin ^{2} a+\left(\sin ^{2} b+\cos ^{2} b\right)+\cos ^{2} c=2$
$\Rightarrow \sin ^{2} a+\cos ^{2} c+1=2$ ....(i)
$\Rightarrow \sin ^{2} a+\cos ^{2} c=1$ ....(ii)
Hence, $R$ is transitive also.
Therefore, relation $R$ is an equivalence relation.