Correct Answer: $\frac{3 \sqrt{3}}{2}$

Solution : Given: $\sec \theta+\tan \theta=\frac{1}{\sqrt{3}}$...................(equation 1)
We know that
$\sec^2\theta-\tan^2\theta=1$
$⇒(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1$
Putting the value, we get
$⇒(\sec\theta-\tan\theta)(\frac{1}{\sqrt3})=1$
$⇒(\sec\theta-\tan\theta)=\sqrt3$............(equation 2)
From equations 1 and 2, we get:
$⇒2\sec\theta=\sqrt3+\frac{1}{\sqrt3}$
$⇒2\sec\theta=\frac{4}{\sqrt3}$
$⇒\sec\theta=\frac{2}{\sqrt3}$
We know that, $\sec\theta=\frac{h}{b}=\frac{2}{\sqrt3}$
So, $p=\sqrt{ h^2-b^2}$
$⇒p=\sqrt{2^2-\sqrt3^2}=1$
$\therefore \cot\theta+\cos\theta=\frac{b}{p}+\frac{b}{h}=\frac{\sqrt3}{1}+\frac{\sqrt3}{2}=\frac{3\sqrt3}{2}$
Hence, the correct answer is $\frac{3\sqrt3}{2}$.