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

Question

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

Accepted Answer

$\begin{gathered} {\sec }^{2}x-{	an }^{2}x=1\, (yx) \ \Rightarrow \ {\sec }^{2}x-{	an }^{2}y=1 \ \Rightarrow 1+{	an }^{2}x+1-{\sec }^{2}y=1 \ \Rightarrow {\sec }^{2}y-{	an }^{2}x=1 \ \Rightarrow symmetric \ {\sec }^{2}x-{	an }^{2}y=1 \ {\sec }^{2}y-{	an }^{2}z=1 \end{gathered}$

Adding both

$\begin{gathered} \Rightarrow {\sec }^{2}x-{	an }^{2}y+{\sec }^{2}y-{	an }^{2}z=1+1 \ {\sec }^{2}x+1-{	an }^{2}z=2 \ {\sec }^{2}x-{	an }^{2}z=1 \ \Rightarrow \end{gathered}$

hence equivalence relation

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