1. Given relation

On the set $S=\left[0,\frac{\pi}{2}\right)$, define $R$ by $xRy \iff \sec^2 x-\tan^2 y=1.$

We use the identity $\sec^2 \theta-\tan^2 \theta=1.$

But here the angles are different, so we simplify carefully.

2. Rewrite the condition

From $\sec^2 x-\tan^2 y=1,$ using $\sec^2 x=1+\tan^2 x$, we get $1+\tan^2 x-\tan^2 y=1.$ So, $\tan^2 x=\tan^2 y.$

Now for $x,y\in \left[0,\frac{\pi}{2}\right)$, we have $\tan x\ge 0,\quad \tan y\ge 0,$ and $\tan t$ is one-one on $\left[0,\frac{\pi}{2}\right)$.

Hence $\tan^2 x=\tan^2 y \implies \tan x=\tan y \implies x=y.$

Therefore the relation is simply $xRy \iff x=y.$ So $R$ is the equality relation on the set $S$.

3. Check reflexive

For every $x\in S$, $\sec^2 x-\tan^2 x=1,$ so $xRx$.

Thus, $R$ is reflexive.

4. Check symmetric

Suppose $xRy$. Then $x=y$. Therefore $y=x$, so $yRx$.

Thus, $R$ is symmetric.

5. Check transitive

Suppose $xRy$ and $yRz$. Then $x=y \quad \text{and} \quad y=z,$ so $x=z$. Hence $xRz$.

Thus, $R$ is transitive.

6. Conclusion

Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation.

So the correct option is $\boxed{\text{D}}.$