If R be a relation defined on (0, π/2) such that xRy ⇒ sec^2x-tan^2y=1, then the relation R is

Question

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

(1) Equivalence relation

(2) Reflexive and transitive only

(3) Symmetric and transitive only

(4) Neither reflexive nor transitive

Answer 1

Correct option is (1) Equivalence relation 

\(x R y \Rightarrow \sec ^2 x-\tan ^2 y=1\)

• \(x R x \Rightarrow \sec ^2 x-\tan ^2 x=1\)

\(\Rightarrow R\) is reflexive

• \(x R y \Rightarrow y R x \)

\(\Rightarrow \sec ^2 x-\tan ^2 y=1 \)

\(\sec ^2 y-\tan ^2 x=\left(1+\tan ^2 y\right)-\left(\sec ^2 x-1\right) \)

\(=2 \sec ^2 x+\tan ^2 y \)

\(=2-\left(\sec ^2 x-\tan ^2 y\right)=2-1=1\)

\(\Rightarrow R\) is symmetric

• \(x R y \Rightarrow y R z\)

\(\Rightarrow \sec ^2 x-\tan ^2 y=1\)

\(\sec ^2 y-\tan ^2 z=1\)

Add \(\Rightarrow \sec ^2 x+\sec ^2 y-\tan ^2 y-\tan _z^2=2\)

\(\Rightarrow \sec ^2 x+(1)-\tan ^2 z=2 \)

\(\Rightarrow \sec ^2 x-\tan ^2 z=1 \)

\(\Rightarrow x R z\)

\(\Rightarrow R\) is transitive.