Question

Let $P$ be the relation defined on the set of all real numbers such that
$P=\left\{\left(a,b\right)\right\}:{\sec }^{2}a-{\tan }^{2}b=1\}$ Then $P$ is :

• 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.

Answer 1

Answer: (D)

for reflexive : ${\sec }^{2}a-{\tan }^{2}a=1$ an identity forall $x\in R\Rightarrow$ reflexive
for symmetric : ${\sec }^{2}a-{\tan }^{2}b=1\dots (i)$ to prove
${\sec }^{2}b-{\tan }^{2}a=1$
${\sec }^{2}b-{\tan }^{2}a=1+{\tan }^{2}b-\left({\sec }^{2}a-1\right)=1$
$+{\tan }^{2}b+1-{\sec }^{2}a={\sec }^{2}a-{\tan }^{2}b=1\Rightarrow$ symmetric
$[\because$ from $(1)]$
for transitive:
${\sec }^{2}a-{\tan }^{2}b=1\dots \dots (ii)$
${\sec }^{2}b-{\tan }^{2}c=1\dots \dots (iii)$
to prove $:se{c}^{2}a-ta{n}^{2}c=1$
proof $L.H.S.$ $1+{\tan }^{2}b+1-{\sec }^{2}b$ from (ii) $\&$ (iii)
$={\sec }^{}b-{\tan }^{2}b$ identify
$=1$
$\Rightarrow P$ is reflexive, symmetric and transitive.