Question

Let P be a relation defined on the set of interval $(0, \frac{\pi}{2}]$ such that P ={(a, b) : $cosec^2a - cot^2b$ = 1}. Then P is

• A. Reflexive and symmetric but not transitive
• B. Reflexive and transitive but not symmetric
• C. Symmetric and transitive but not reflexive
• D. Equivalence relation

Answer 1

Answer: (D)

An equivalence relation

Answer 2

We are given the relation

$$P = \{(a,b) : \csc^2 a - \cot^2 b = 1\}$$

on the set $(0, \tfrac{\pi}{2}]$.

Step 1: Check Reflexivity

For $(a,a) \in P$, we have:

$$\csc^2 a - \cot^2 a = 1$$

But using the identity $\csc^2 a = 1 + \cot^2 a$, we get:

$$(1 + \cot^2 a) - \cot^2 a = 1.$$

Thus, the relation is reflexive.

Step 2: Analyze the Given Equation

If $(a,b) \in P$, then

$$\csc^2 a - \cot^2 b = 1 \quad \Longrightarrow \quad \csc^2 a = 1 + \cot^2 b.$$

But also by the trigonometric identity, we have for any $a$:

$$\csc^2 a = 1 + \cot^2 a.$$

Comparing these,

$$1 + \cot^2 a = 1 + \cot^2 b \quad \Longrightarrow \quad \cot^2 a = \cot^2 b.$$

Since $a, b \in (0, \tfrac{\pi}{2}]$, $\cot$ is positive; hence,

$$\cot a = \cot b \quad \Longrightarrow \quad a = b.$$

Step 3: Conclude the Relation Properties

The relation $P$ contains only pairs $(a,b)$ where $a = b$. This means:

• It is symmetric (if $a=b$ then automatically $b=a$).
• It is transitive (if $a=b$ and $b=c$, then $a=c$).

Thus, $P$ is an equivalence relation.