Correct Answer: $\frac{\sqrt{b^2-a^2}}{a}$

Solution : Given,
$\operatorname{cosec} \theta=\frac{b}{a}$
⇒ $\sin\theta=\frac ab$
We have to find the value of $\frac{\sqrt{3} \cot \theta+1}{\tan \theta+\sqrt{3}}$
We know, $\cot\theta=\frac{\cos\theta}{\sin\theta}$ and $\tan\theta=\frac{\sin\theta}{\cos\theta}$
$\frac{\sqrt{3} \cot \theta+1}{\tan \theta+\sqrt{3}}=\frac{\sqrt3\times\frac{\cos\theta}{\sin\theta}+1}{\frac{\sin\theta}{\cos\theta}+\sqrt3}$
= $\frac{\cos\theta(\sqrt3\cos\theta+\sin\theta)}{\sin\theta(\sin\theta+\sqrt3\cos\theta)}$
= $\frac{\cos\theta}{\sin\theta}$
= $\frac{\sqrt{1-\sin^2\theta}}{\sin\theta}$ [As $\sin^2\theta+\cos^2\theta=1$]
= $\frac{\sqrt{1-(\frac ab)^2}}{\frac ab}$
= $\frac{\sqrt{b^2-a^2}}{a}$
Hence, the correct answer is $\frac{\sqrt{b^2-a^2}}{a}$.