If R be a relation defined on \((0,\pi /2)\) such that \(xRy\Rightarrow {\sec }^{2}x-{\tan }^{2}y=1\), then the relation…

\(xRy\Rightarrow {\sec }^{2}x-{\tan }^{2}y=1\\ xRx\Rightarrow {\sec }^{2}x-{\tan }^{2}x=1\\ \Rightarrow R\text{ is reflexive }\\ xRy\Rightarrow yRx\\ \Rightarrow {\sec }^{2}x-{\tan }^{2}y=1\\ {\sec }^{2}y-{\tan }^{2}x=\left(1+{\tan }^{2}y\right)-\left({\sec }^{2}x-1\right)\\ =2-{\sec }^{2}x+{\tan }^{2}y\\ =2-\left({\sec }^{2}x-{\tan }^{2}y\right)=2-1=1\\ \Rightarrow R\text{ is symmetric }\\ xRy\Rightarrow yRz\\ \Rightarrow {\sec }^{2}x-{\tan }^{2}y=1\\ {\sec }^{2}y-{\tan }^{2}z=1\\ \text{ Add }\Rightarrow {\sec }^{2}x+{\sec }^{2}y-{\tan }^{2}y-{\tan }^{2}z=2\\ \Rightarrow {\sec }^{2}x+(1)-{\tan }^{2}z=2\\ \Rightarrow xRz\\ \Rightarrow R\text{ is transitive. }\\\)