Trigonometric Identities — Maths STD 10 — Question

Given, $\text{x}=\text{a}\sec\theta,\text{y}=\text{b}\tan\theta$
So,
$\text{b}^2\text{y}^2-\text{a}^2\text{y}^2$
$=\text{b}^2(\text{a}\sec\theta)^2-\text{a}^2(\text{b}\tan\theta)^2$
$=\text{b}^2\text{a}^2\sec^2\theta-\text{a}^2\text{b}^2\tan^2\theta$
$=\text{b}^2\text{a}^2(\sec^2\theta-\tan^2\theta)$
We know that,
$\sec^2\theta-\tan^2\theta=1$
$\text{b}^2\text{x}^2-\text{a}^2\text{y}^2=\text{a}^2\text{b}^2$
Hence, the correct option is $(D)$.