Question

For any two real numbers $\theta$ and $\phi$ we define $\theta R \phi$ if and only if $\sec ^{2} \theta-\tan ^{2} \phi=1$. The relation $R$ is

• A. Reflexive but not transitive
• B. Symmetric but not reflexive
• C. Both reflexive and symmetric but not transitive
• D. An equivalence relation

Answer 1

Answer: (D)

Given relation is defined as
$\theta R \phi$ such that $\sec ^{2} \theta-\tan ^{2} \phi=1$
For Reflexive
When $\theta R \theta$
$\sec ^{2} \theta-\tan ^{2} \theta=1$
$\Rightarrow 1=1$, which is true.
Thus, it is reflexive.
For Symmetric
When $\theta R \phi$
$\sec ^{2} \theta-\tan ^{2} \phi=1$
$\Rightarrow \left(1+\tan ^{2} \theta\right)-\left(\sec ^{2} \phi-1\right)=1$
$\Rightarrow 2+\tan ^{2} \theta-\sec ^{2} \phi=1$
$\Rightarrow \sec ^{2} \phi-\tan ^{2} \theta=1$
$\Rightarrow \phi R \theta$
Thus, it is symmetric.
For Transitive
When $\theta R \phi$ and $\phi R \psi$, then
$\sec ^{2} \theta-\tan ^{2} \phi=1$
and $\sec ^{2} \phi-\tan ^{2} \psi=1$
Now, $\theta R \psi$
Then, $\sec ^{2} \theta-\tan ^{2} \psi=1$
$\Rightarrow \sec ^{2} \theta-\tan ^{2} \psi+1=1+1$
$\Rightarrow \sec ^{2} \theta-\tan ^{2} \psi+\sec ^{2} \phi-\tan ^{2} \phi=$
$\Rightarrow \theta R \phi$ and $\phi R \psi$
Thus, it is transitive.
Hence, it is an equivalence relation.