Define a relation R on the interval [0, π/2) by xRy if and only if sec²x - tan²y = 1. Then R is:
(1)

Q: Wait, why do we reject the negative root when taking $\sqrt{\tan^{2}x} = \sqrt{\tan^{2}y}$?

Great catch! In the interval $[0, \frac{\pi}{2})$, the tangent function is always non-negative. Since $\tan x \geq 0$ and $\tan y \geq 0$, the equation $\tan x = -\tan y$ would imply $\tan x = \tan y = 0$, which only happens at $x=y=0$. For all other values, the negative root is impossible because the tangent function is strictly positive in the first quadrant.

Q: Why is the interval open at $\frac{\pi}{2}$?

Because $\tan(\frac{\pi}{2})$ is undefined! The relation is defined on a valid domain where the tangent function exists. Always check the boundaries of your domain before jumping into calculations; it is a classic JEE trap.

Q: Is the identity relation always an equivalence relation?

Absolutely! The identity relation $xRy \iff x=y$ is the 'gold standard' of equivalence relations. It is reflexive ($x=x$), symmetric (if $x=y$, then $y=x$), and transitive (if $x=y$ and $y=z$, then $x=z$). It satisfies all three conditions perfectly.

Question

Define a relation R on the interval $[0,\frac{\pi}{2})$ by xRy if and only if $sec^{2}x-tan^{2}y=1$. Then R is:

Options:

(A) both reflexive and transitive but not symmetric

(B) an equivalence relation

(C) reflexive but neither symmetric not transitive

(D) both reflexive and symmetric but not transitive

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, aspiring engineers! Today, we are going to unravel a beautiful problem that sits at the intersection of trigonometry and set theory. Imagine you are standing on a coordinate plane, looking at the interval $[0, \frac{\pi}{2})$.

We are given a relation $R$ defined by the condition:
$$\sec^{2}x - 	an^{2}y = 1$$

At first glance, this might look like a daunting, abstract equation, but let's break it down with the precision of a master architect.

### The Identity Bridge

The secret to solving this lies in the fundamental trigonometric identity:
$$\sec^{2}x - 	an^{2}x = 1$$

By rearranging this, we get $\sec^{2}x = 1 + 	an^{2}x$. This is our bridge!

By substituting this into our original relation, we transform the equation into:
$$(1 + 	an^{2}x) - 	an^{2}y = 1$$

Suddenly, the complexity melts away. Subtracting $1$ from both sides, we are left with the elegant result:
$$	an^{2}x = 	an^{2}y$$

### The Domain Trap

Now, here is where many students stumble. We have $	an^{2}x = 	an^{2}y$.

If we were on the entire real number line, we might be tempted to say $	an x = \pm 	an y$. But look at our domain: $[0, \frac{\pi}{2})$.

In this interval, the tangent function is always non-negative. Because $	an x \geq 0$ and $	an y \geq 0$, the negative root is physically impossible. Thus, we are left with the clean, simple condition:
$$	an x = 	an y$$

### The Injective Insight

Why does $	an x = 	an y$ imply $x = y$? Because the tangent function is strictly increasing on the interval $[0, \frac{\pi}{2})$.

In mathematical terms, it is a one-to-one or injective function. It never repeats a value. Therefore, if the outputs are equal, the inputs must be equal.

We have arrived at the identity relation:
$$xRy \iff x = y$$

### The Equivalence Verdict

Now, let's test the three pillars of an equivalence relation.

First, is it reflexive? Yes, because $x = x$ is always true.

Second, is it symmetric? Yes, because if $x = y$, then $y = x$.

Finally, is it transitive? Yes, because if $x = y$ and $y = z$, then $x = z$.

Since our relation satisfies all three properties, it is, by definition, an **equivalence relation**. You have just navigated a complex problem by using fundamental identities and domain constraints. Keep this logical clarity, and you will conquer any JEE problem!

Animated Solution for Mathematics - Sets and Relations: Define a relation R on the interval $[0, \frac{π}{2})$ by xRy if and only if $se c^{2} x - t a n^{2} y = 1$ . Then R is:

Select Answer:

Visualized Solution

Understanding the Relation $R$

Recalling the Trigonometric Identity

Substituting the Identity

Simplifying the Equation

Analyzing the Domain Constraints

The One-to-One Property

The Identity Relation

Checking Reflexivity

Checking Symmetry

Checking Transitivity

Conclusion: Equivalence Relation

The Sigma Insight: Types of Relations

Analyzing the Setup

The Identity Bridge

The Domain Trap

The Injective Insight

The Equivalence Verdict

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Animated Solution for Mathematics - Sets and Relations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Define a relation R on the interval [0,2π​) by xRy if and only if sec2x−tan2y=1. Then R is: