Define a relation R on the interval $ \left[0, \frac{\pi}{2}\right) $ by $ x $ R $ y $ if and only if $ \sec^2x - \tan^2y = 1 $. Then R is :
Solution
<p>$$\begin{aligned}
& \sec ^2 x-\tan ^2 x=1 \quad(\text { on replacing } y \text { with } x) \\
& \Rightarrow \text { Reflexive } \\
& \sec ^2 x-\tan ^2 y=1 \\
& \Rightarrow 1+\tan ^2 x+1-\sec ^2 y=1 \\
& \Rightarrow \sec ^2 y-\tan ^2 x=1 \\
& \Rightarrow \operatorname{symmetric} \\
& \sec ^2 x-\tan ^2 y=1 \\
& \sec ^2 y-\tan ^2 z=1
\end{aligned}$$</p>
<p>Adding both</p>
<p>$$\begin{aligned}
& \Rightarrow \sec ^2 x-\tan ^2 y+\sec ^2 y-\tan ^2 z=1+1 \\
& \sec ^2 x+1-\tan ^2 z=2 \\
& \sec ^2 x-\tan ^2 z=1 \\
& \Rightarrow \text { Transitive }
\end{aligned}$$</p>
<p>hence equivalence relation</p>
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
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