Define a relation R on the interval [0,2) by xRy if and only if ^2x - ^2y = 1.

Question

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

Accepted Answer

$$\begin{aligned}
& \sec ^2 x-\tan ^2 x=1 \quad(\text { on replacing } y \text { with } x) \
& \Rightarrow \text { Reflexive } \
& \sec ^2 x-\tan ^2 y=1 \
& \Rightarrow 1+\tan ^2 x+1-\sec ^2 y=1 \
& \Rightarrow \sec ^2 y-\tan ^2 x=1 \
& \Rightarrow \operatorname{symmetric} \
& \sec ^2 x-\tan ^2 y=1 \
& \sec ^2 y-\tan ^2 z=1
\end{aligned}$$

Adding both

$$\begin{aligned}
& \Rightarrow \sec ^2 x-\tan ^2 y+\sec ^2 y-\tan ^2 z=1+1 \
& \sec ^2 x+1-\tan ^2 z=2 \
& \sec ^2 x-\tan ^2 z=1 \
& \Rightarrow \text { Transitive }
\end{aligned}$$

hence equivalence relation

Solutions · 1