$\sec \theta+\tan \theta=\sqrt{3}$

$\Rightarrow \quad 1+\sin \theta=\sqrt{3} \cos \theta(\cos \theta \neq 0)$

$\Rightarrow \quad \sqrt{3} \cos \theta-\sin \theta=1$

$\Rightarrow \quad \frac{\sqrt{3}}{2} \cos \theta-\frac{1}{2} \sin \theta=\frac{1}{2}$

$\Rightarrow \quad \cos \theta \cos \frac{\pi}{6}-\sin \theta \sin \frac{\pi}{6}=\cos \frac{\pi}{3}$

$\Rightarrow \quad \cos \left(\theta+\frac{\pi}{6}\right)=\cos \frac{\pi}{3}$

$\Rightarrow \quad \theta+\frac{\pi}{6}=2 n \pi \pm \frac{\pi}{3}$             $\left( {n \in Z} \right)$

$\Rightarrow \quad \theta=2 n \pi-\frac{\pi}{6} \pm \frac{\pi}{3}$              $\left( {n \in Z} \right)$

$=2 n \pi+\frac{\pi}{6} \text { or } 2 n \pi-\frac{\pi}{2}$

But we reject the values $\theta=2 \mathrm{n} \pi-\frac{\pi}{2}$

$\because \quad \cos \theta=0$ for this values of $\theta$

$\therefore $ $\theta=2 \mathrm{n} \pi+\frac{\pi}{6}, \mathrm{n} \in \mathrm{I}$

Thus, $\theta=\frac{\pi}{6}$ is the only value of $\theta \in[0,2, \pi]$