Relations and Functions – Types of Relations – JEE Main 29 Jan 2025 Shift 1

Solution:

Given relation: $x R y \iff \sec^{2}x – \tan^{2}y = 1$.

**Check Reflexivity:**

For any $x \in [0, \frac{\pi}{2}]$, we check if $x R x$.

$\sec^{2}x – \tan^{2}x = 1$.

This is a standard trigonometric identity, so it is always true. Thus, $R$ is reflexive.

**Check Symmetry:**

Let $x R y$, so $\sec^{2}x – \tan^{2}y = 1$.

We know $\sec^{2}\theta = 1 + \tan^{2}\theta$. Substitute this for $x$:

$(1 + \tan^{2}x) – \tan^{2}y = 1$

$\tan^{2}x – \tan^{2}y = 0 \Rightarrow \tan^{2}x = \tan^{2}y$.

Now check $y R x$:

$\sec^{2}y – \tan^{2}x = (1 + \tan^{2}y) – \tan^{2}x$.

Since $\tan^{2}y = \tan^{2}x$, this becomes $(1 + \tan^{2}x) – \tan^{2}x = 1$.

Thus, $y R x$ is true. $R$ is symmetric.

**Check Transitivity:**

Let $x R y$ and $y R z$.

(1) $\sec^{2}x – \tan^{2}y = 1$

(2) $\sec^{2}y – \tan^{2}z = 1$

Adding (1) and (2):

$\sec^{2}x – \tan^{2}y + \sec^{2}y – \tan^{2}z = 2$

$\sec^{2}x + (\sec^{2}y – \tan^{2}y) – \tan^{2}z = 2$

Since $\sec^{2}y – \tan^{2}y = 1$:

$\sec^{2}x + 1 – \tan^{2}z = 2$

$\sec^{2}x – \tan^{2}z = 1$

Thus, $x R z$. $R$ is transitive.

Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation.

Ans. (1)