Question

For any two real numbers $a$ and $b$, we define $a$ $R$ $b$ of 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

Let the given relation defined as
$R=\left\{\right(a,b)|{\sin }^{2}a+{\cos }^{2}a=1\}$
For reflexive, ${\sin }^{2}a+{\cos }^{2}a=1$ $(\because {\sin }^2\theta +{\cos }^2\theta =1,forall\theta \in R)$
$\Rightarrow {}_a{R}_a\Rightarrow (a,a)\in R$
$\Rightarrow$ $R$ is reflexive
Foe 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 $aRb,bRc$
$\Rightarrow {\sin }^{2}a+{\cos }^{2}b=\mathrm{1...}\left(i\right)$
and ${\sin }^{2}b+{\cos }^{2}c=\mathrm{1....}\left(ii\right)$
On adding eqs. $\left(i\right)$ and $\left(ii\right)$ we get
${\sin }^{2}a+({\sin }^{2}b+{\cos }^{2}b)+{\cos }^{2}c=2$
$\Rightarrow {\sin }^{2}a+{\cos }^{2}c+1=2$
$\Rightarrow {\sin }^{2}a+{\cos }^{2}c=1$
Hence, $R$ is transitive also
Therefore relation $R$ is an equivalence relation.