Mathematics - Sets and Relations Question with Solution | TestHub

$\begin{array}{rl}& {\sec }^{2}x-{\tan }^{2}x=1(\text{ on replacing }y\text{ with }x)\\ & \Rightarrow \text{ Reflexive }\\ & {\sec }^{2}x-{\tan }^{2}y=1\\ & \Rightarrow 1+{\tan }^{2}x+1-{\sec }^{2}y=1\\ & \Rightarrow {\sec }^{2}y-{\tan }^{2}x=1\\ & \Rightarrow \text{ symmetric }\\ & {\sec }^{2}x-{\tan }^{2}y=1\\ & {\sec }^{2}y-{\tan }^{2}z=1\end{array}$ Adding both $\begin{array}{rl}& \Rightarrow {\sec }^{2}x-{\tan }^{2}y+{\sec }^{2}y-{\tan }^{2}z=1+1\\ & {\sec }^{2}x+1-{\tan }^{2}z=2\\ & {\sec }^{2}x-{\tan }^{2}z=1\\ & \Rightarrow \text{ Transitive }\end{array}$ hence equivalence releation