A) reflexive and symmetric but not transitive.
B) reflexive and transitive but not symmetric.
C) symmetric and transitive but not reflexive.
D) an equivalence relation.
Correct Answer: D
Solution :
\[P=\{(a,b):se{{c}^{2}}a-ta{{n}^{2}}b=1\}\] For reflexive : \[{{\sec }^{2}}a-{{\tan }^{2}}a=1\] \[(true\,\forall a)\] For symmetric: \[{{\sec }^{2}}b-{{\tan }^{2}}a=1\] L.H.S \[1+{{\tan }^{2}}b-(se{{c}^{2}}a-1)=1+ta{{n}^{2}}b-{{\sec }^{2}}a+1\] \[=-(se{{c}^{2}}a-ta{{n}^{2}}b)+2\] \[=-1+2=1\] So, Relation is symmetric For transitive : if \[{{\sec }^{2}}a-{{\tan }^{2}}b=1\]and\[{{\sec }^{2}}b-{{\tan }^{2}}c=1\] \[{{\sec }^{2}}a-{{\tan }^{2}}c=(1+ta{{n}^{2}}b)-(se{{c}^{2}}b-1)\] \[=-{{\sec }^{2}}b+{{\tan }^{2}}b+2\] \[=-1+2=1\]So, Relation is transitive. Hence, Relation P is an equivalence relation
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