JEE Main 2014 — Mathematics Algebra

$$P={(a,b):{\mathrm{sec}}^{2}a-{\mathrm{tan}}^{2}b=1}$$P=(a,b):sec2a−tan2b=1

$$P={(a,a):{\mathrm{sec}}^{2}a-{\mathrm{tan}}^{2}a=1}$$P=(a,a):sec2a−tan2a=1

its reflexive relation.

Now $P={(a,b):{\mathrm{sec}}^{2}a-{\mathrm{tan}}^{2}b=1}$P=(a,b):sec2a−tan2b=1

$${\mathrm{sec}}^{2}a-{\mathrm{tan}}^{2}b=1$$sec2a−tan2b=1

$$\Rightarrow {\mathrm{sec}}^{2}a=1+{\mathrm{tan}}^{2}b$$⇒sec2a=1+tan2b

$$\Rightarrow {\mathrm{sec}}^{2}a={\mathrm{sec}}^{2}b$$⇒sec2a=sec2b

$$\Rightarrow |\mathrm{sec}a|=|\mathrm{sec}b|$$⇒∣seca∣=∣secb∣

$${\mathrm{sec}}^{2}a-{\mathrm{tan}}^{2}b=1$$sec2a−tan2b=1

$$1+{\mathrm{tan}}^{2}a-{\mathrm{sec}}^{2}b+1=1$$1+tan2a−sec2b+1=1

$${\mathrm{sec}}^{2}b-{\mathrm{tan}}^{2}a=1$$sec2b−tan2a=1

Hence, it is symmetric.

If $|\mathrm{sec}a|=|\mathrm{sec}b|$∣seca∣=∣secb∣ and $|\mathrm{sec}b|=|\mathrm{sec}c|$∣secb∣=∣secc∣ then $|\mathrm{sec}a|=|\mathrm{sec}c|\Rightarrow$∣seca∣=∣secc∣⇒ transitive

So, it is an equivalence relation.