Question

For any two real numbers $\theta$ and $\varphi$ we define $\theta R\varphi$ if and only if ${\sec }^2\theta -{\tan }^2\varphi =1$. 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
$\theta R\varphi$ such that ${\sec }^2\theta -{\tan }^2\varphi =1$
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 $\theta R\varphi$
${\sec }^2\theta -{\tan }^2\varphi =1$
$\Rightarrow \left(1+{\tan }^2\theta \right)-\left({\sec }^2\varphi -1\right)=1$
$\Rightarrow 2+{\tan }^2\theta -{\sec }^2\varphi =1$
$\Rightarrow {\sec }^2\varphi -{\tan }^2\theta =1$
$\Rightarrow \varphi R\theta$
Thus, it is symmetric.
For Transitive
When $\theta R\varphi$ and $\varphi R\psi$, then
${\sec }^2\theta -{\tan }^2\varphi =1$
and ${\sec }^2\varphi -{\tan }^2\psi =1$
Now, $\theta R\psi$
Then, ${\sec }^2\theta -{\tan }^2\psi =1$
$\Rightarrow {\sec }^2\theta -{\tan }^2\psi +1=1+1$
$\Rightarrow {\sec }^2\theta -{\tan }^2\psi +{\sec }^2\varphi -{\tan }^2\varphi =$
$\Rightarrow \theta R\varphi$ and $\varphi R\psi$
Thus, it is transitive.
Hence, it is an equivalence relation.