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 : $\sec ^{2} a-\tan ^{2} b=1 \ldots(i)$ 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:
$\sec ^{2} a-\tan ^{2} b=1 \ldots \ldots(i i)$
$\sec ^{2} b-\tan ^{2} c=1 \ldots \ldots(i i i)$
to prove $: sec ^{2} a-tan ^{2} c=1$
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.