Question

For any two real number $\theta$ and $\varphi$ where, $\theta ,\varphi \in (\frac{-\pi }{2},\frac{\pi }{2}).$ We define $\theta R\varphi$ if and only if ${\sec }^2\theta -{\tan }^2\varphi =1.$ Then relation $R$ is

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

Answer 1

The correct option is D An equivalence relation

$\because {\sec }^2\theta -{\tan }^2\theta =1\forall \theta \in (\frac{-\pi }{2},\frac{\pi }{2})$ So, $R$ is a reflexive relation

Let $(\theta ,\varphi )\in R\Rightarrow {\sec }^2\theta -{\tan }^2\varphi =1$
$\Rightarrow 1+{\tan }^2\theta -({\sec }^2\varphi -1)=1$
$\Rightarrow {\sec }^2\varphi -{\tan }^2\theta =1$
$\Rightarrow (\varphi ,\theta )\in R$
So, $R$ is a symmetric relation.

Let $(\theta ,\varphi )\in R\Rightarrow {\sec }^2\theta -{\tan }^2\varphi =1\dots \left(1\right)$
and $(\varphi ,\alpha )\in R\Rightarrow {\sec }^2\varphi -{\tan }^2\alpha =1\dots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$
$\Rightarrow {\sec }^2\theta -{\tan }^2\alpha =1$
$\Rightarrow (\theta ,\alpha )\in R$
So, $R$ is a transitive relation.
Hence, $R$ is an equivalence relation