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

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Aicharya · Accepted Answer

$\begin{aligned} & \sec ^2 x-	an ^2 x=1 \quad(	ext { on replacing } y 	ext { with } x) \ & \Rightarrow 	ext { Reflexive } \ & \sec ^2 x-	an ^2 y=1 \ & \Rightarrow 1+	an ^2 x+1-\sec ^2 y=1 \ & \Rightarrow \sec ^2 y-	an ^2 x=1 \ & \Rightarrow 	ext { symmetric } \ & \sec ^2 x-	an ^2 y=1 \ & \sec ^2 y-	an ^2 z=1 \end{aligned}$ Adding both $\begin{aligned} & \Rightarrow \sec ^2 x-	an ^2 y+\sec ^2 y-	an ^2 z=1+1 \ & \sec ^2 x+1-	an ^2 z=2 \ & \sec ^2 x-	an ^2 z=1 \ & \Rightarrow 	ext { Transitive } \end{aligned}$ hence equivalence releation *

Define a relation R on the interval \(\left[0, \frac{\pi}{2}\right)\) by \(x \mathrm{R} y\) if and only if…

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