If $2{\cos }^2\theta -5\cos \theta +2=0$, $0^{\circ }<\theta <90^{\circ }$, then the value of $\frac{1}{\csc \theta +\cot \theta }$ is:

Given quadratic equation in terms of $\cos \theta$:
$$2{\cos }^2\theta -5\cos \theta +2=0$$
Let $\cos \theta =y$:
$$2y^2-5y+2=0$$
$$2y^2-4y-y+2=0$$
$$2y(y-2)-1(y-2)=0$$
$$(2y-1)(y-2)=0$$
So, $y=2$ or $y=\frac{1}{2}$.Since $\cos \theta \le 1$, $y=2$ is rejected. Thus, $\cos \theta =\frac{1}{2}$.Since $0^{\circ }<\theta <90^{\circ }$, $\theta =60^{\circ }$.Now substitute $\theta =60^{\circ }$ into the target expression:
$$\csc 60^{\circ }=\frac{2}{\sqrt{3}}$$
$$\cot 60^{\circ }=\frac{1}{\sqrt{3}}$$
$$\csc 60^{\circ }+\cot 60^{\circ }=\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}$$
Thus, the value of the required expression is:
$$\frac{1}{\csc \theta +\cot \theta }=\frac{1}{\sqrt{3}}$$