Define a relation R on the interval by xRy if and only if . Then R is : [2025]

Question

Define a relation R on the interval $[0, \frac{π}{2})$ by xRy if and only if $\sec^{2} x - \tan^{2} y = 1$ . Then R is : [2025]

Smart Achievers · Accepted Answer

(1)

$[0, \frac{π}{2})$

$∴$ R is reflexive

Consider, $\sec^{2} x - \tan^{2} y = 1$

$\Rightarrow 1 + \tan^{2} x - (\sec^{2} y - 1) = 1$

$\Rightarrow 1 + \tan^{2} x - s e c^{2} y + 1 = 1 \Rightarrow \sec^{2} y - \tan^{2} x = 1$

$∴$ R is symmetric.

Now, if $\sec^{2} x - \tan^{2} y = 1$ and $\sec^{2} y - \tan^{2} z = 1$

Adding both equation, $\sec^{2} x - \tan^{2} y + \sec^{2} y - \tan^{2} z = 2$

$\Rightarrow \sec^{2} x - \tan^{2} z = 1$ [ $∵ \sec^{2} y - \tan^{2} y = 1$ ]

$∴$ R is transitive

Thus R is an equivalence relation.

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