Mathematics - Sets and Relations Question with Solution | TestHub

$P=\left\{\left(a,b\right):{\sec }^{2}a-{\tan }^{2}b=1\right\}$

$P=\left\{\left(a,a\right):{\sec }^{2}a-{\tan }^{2}a=1\right\}$

its reflexive relation.

Now $P=\left\{\left(a,b\right):{\sec }^{2}a-{\tan }^{2}b=1\right\}$

${\sec }^{2}a-{\tan }^{2}b=1$

$\Rightarrow {\sec }^{2}a=1+{\tan }^{2}b$

$\Rightarrow {\sec }^{2}a={\sec }^{2}b$

$\Rightarrow \left|\sec a\right|=\left|\sec b\right|$

${\sec }^{2}a-{\tan }^{2}b=1$

$1+{\tan }^{2}a-{\sec }^{2}b+1=1$

${\sec }^{2}b-{\tan }^{2}a=1$

Hence, it is symmetric.

If $\left|\sec a\right|=\left|\sec b\right|$ and $\left|\sec b\right|=\left|\sec c\right|$ then $\left|\sec \mathrm{a}\right|=\left|\sec c\right|\Rightarrow$ transitive 

So, it is an equivalence relation.