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

The correct option is D An equivalence relation

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 $\theta R\varphi$
${\sec }^2\theta -{\tan }^2\varphi =1$
$\Rightarrow (1+{\tan }^2\theta )-({\sec }^2\varphi -1)=1$
$\Rightarrow 2+{\tan }^2\theta -{\sec }^2\varphi =1$
$\Rightarrow {\sec }^2\varphi -{\tan }^2\theta =1$
$\Rightarrow \varphi R\theta$
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, ${\sec }^2\theta -{\tan }^2\psi =1$
$\Rightarrow {\sec }^2\theta -{\tan }^2\psi +1=1+1$
$\Rightarrow {\sec }^2\theta -{\tan }^2\psi +{\sec }^2\varphi -{\tan }^2\varphi =1+1$
$\Rightarrow \theta R\varphi$ and $\varphi R\psi$
Thus, it is transitive.

Hence, it is an equivalence relation.