For given hyperbola the eccentricity is given as

$e^2=1+\frac{{\mathrm{s}\mathrm{i}\mathrm{n}}^2\theta }{{\mathrm{c}\mathrm{o}\mathrm{s}}^2\theta }=1+{\mathrm{t}\mathrm{a}\mathrm{n}}^2\theta {=\mathrm{s}\mathrm{e}\mathrm{c}}^2\theta$

$\Rightarrow e=\mathrm{s}\mathrm{e}\mathrm{c}\theta$

$\therefore$ Length of latus rectum $l=\frac{2{\mathrm{s}\mathrm{i}\mathrm{n}}^2\theta }{\mathrm{c}\mathrm{o}\mathrm{s}\theta }=\frac{2{\mathrm{t}\mathrm{a}\mathrm{n}}^2\theta }{\mathrm{s}\mathrm{e}\mathrm{c}\theta }$

$\Rightarrow l=\frac{2(e^2-1)}{e}=2\left(e-\frac{1}{e}\right)$
On differentiating w.r.t. $e$, we get

$\frac{dl}{de}=2\left(1+\frac{1}{e^2}\right)>0$

$\therefore$ $l$ is an increasing function

$\therefore l_{\mathrm{m}\mathrm{i}\mathrm{n}}=2\left(2-\frac{1}{2}\right)=3$

$\therefore$ Range of latus rectum is $\left(3,\infty \right)$.