Question

For any two real numbers $a$ and $b$, we define $aRb$ if and only if $si{n}^{2}a+co{s}^{2}b=1$. 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

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