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

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