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1.Relation and Function
medium
Define a relation $R$ on the interval $\left[0, \frac{\pi}{2}\right)$ by $xRy$ if and only if $\sec ^2 x-\tan ^2 y=1$. Then $R$ is :
Aan equivalence relation
Bboth reflexive and transitive but not symmetric
Cboth reflexive and symmetric but not transitive
Dreflexive but neither symmetric not transitive
(JEE MAIN-2025)
Solution
$\sec ^2 x-\tan ^2 x=1 \text { (on replacing } y \text { with } x \text { ) }$
$\Rightarrow \text { Reflexive }$
$\sec ^2 x-\tan ^2 y=1$
$\Rightarrow 1+\tan ^2 x+1-\sec ^2 y=1$
$\Rightarrow \sec ^2 y-\tan ^2 x=1$
$\Rightarrow \text { symmetric }$
$\sec ^2 x-\tan ^2 y=1$
$\sec ^2 y-\tan ^2 z=1$
Adding both
$\Rightarrow \sec ^2 x-\tan ^2 y+\sec ^2 y-\tan ^2 z=1+1$
$\sec ^2 x+1-\tan ^2 z=2$
$\sec ^2 x-\tan ^2 z=1$
$\Rightarrow \text { Transitive }$
hence equivalence releation
$\Rightarrow \text { Reflexive }$
$\sec ^2 x-\tan ^2 y=1$
$\Rightarrow 1+\tan ^2 x+1-\sec ^2 y=1$
$\Rightarrow \sec ^2 y-\tan ^2 x=1$
$\Rightarrow \text { symmetric }$
$\sec ^2 x-\tan ^2 y=1$
$\sec ^2 y-\tan ^2 z=1$
Adding both
$\Rightarrow \sec ^2 x-\tan ^2 y+\sec ^2 y-\tan ^2 z=1+1$
$\sec ^2 x+1-\tan ^2 z=2$
$\sec ^2 x-\tan ^2 z=1$
$\Rightarrow \text { Transitive }$
hence equivalence releation
Standard 12
Mathematics
