Find the value of $\csc \theta (1-\cos \theta )(\csc \theta +\cot \theta )$.

Given expression:
$$\csc \theta (1-\cos \theta )(\csc \theta +\cot \theta )$$

Let us multiply $\csc \theta$ into the first parentheses $(1-\cos \theta )$:

$$\csc \theta (1-\cos \theta )=\csc \theta -\csc \theta \cdot \cos \theta$$

We know that $\csc \theta =\frac{1}{\sin \theta }$, so:

$$\csc \theta \cdot \cos \theta =\frac{1}{\sin \theta }\cdot \cos \theta =\frac{\cos \theta }{\sin \theta }=\cot \theta$$

Thus, the first part simplifies to:

$$\csc \theta -\cot \theta$$

Now substitute this back into the original complete expression:

$$(\csc \theta -\cot \theta )(\csc \theta +\cot \theta )$$

Using the algebraic identity $(a-b)(a+b)=a^2-b^2$:

$$={\csc }^2\theta -{\cot }^2\theta$$

From the standard fundamental identity of trigonometry, we know that:

$${\csc }^2\theta -{\cot }^2\theta =1$$

Therefore, the value of the expression is $1$.