Question

For any two real numbers $\theta$ and $\varphi$, we define $\theta R\varphi$, if and only if $se{c}^2\theta –{\tan }^2\varphi =1$.

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

Explanation for the correct option:

Step 1. For Reflexive:

Given relation is defined as

$\theta R\varphi$ such that $se{c}^2\theta -{\tan }^2\varphi =1$

When $\theta R\theta$

$se{c}^2\theta -{\tan }^2\theta =1$

$\Rightarrow$ $1=1$, which is true.

Thus, it is reflexive.

Step 2. For Symmetric:

When $\theta R\varphi$

$se{c}^2\theta -{\tan }^2\varphi =1$

$\Rightarrow$$(1+{\tan }^2\theta )-(se{c}^2\varphi -1)=1$

$\Rightarrow$ $2+{\tan }^2\theta -se{c}^2\varphi =1$

$\Rightarrow$ $se{c}^2\varphi -{\tan }^2\theta =1$

$\Rightarrow$ $\varphi R\theta$

Thus, it is symmetric.

Step 3. For Transitive:

When $\theta R\varphi$ and $\varphi R\psi$, then

$se{c}^2\theta -{\tan }^2\varphi =1$

and $se{c}^2\varphi -{\tan }^2\psi =1$

Now, $\theta R\psi$

Then, $se{c}^2\theta -{\tan }^2\psi =1$

$\Rightarrow$ $se{c}^2\theta -{\tan }^2\psi +1=1+1$

$\Rightarrow$$se{c}^2\theta -{\tan }^2\psi +se{c}^2\varphi -{\tan }^2\varphi =1+1$

$\Rightarrow$$\theta R\varphi and\varphi R\psi$

Thus, it is transitive.

$\therefore$It is an equivalence relation

Hence, Option ‘D’ is Correct.