If $\cot \theta =\frac{15}{8}$, then what is the value of $\frac{(2+2\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(2-2\cos \theta )}$?

Let us simplify the given expression algebraically:
$$\text{Expression}=\frac{2(1+\sin \theta )(1-\sin \theta )}{2(1+\cos \theta )(1-\cos \theta )}$$

Canceling out the common factor of 2 and applying the difference of squares identity $(a+b)(a-b)=a^2-b^2$:

$$\text{Expression}=\frac{1-{\sin }^2\theta }{1-{\cos }^2\theta }$$

From basic trigonometric identities, we know $1-{\sin }^2\theta ={\cos }^2\theta$ and $1-{\cos }^2\theta ={\sin }^2\theta$. Substituting these into the fraction:

$$\text{Expression}=\frac{{\cos }^2\theta }{{\sin }^2\theta }={\cot }^2\theta$$

Given that $\cot \theta =\frac{15}{8}$, squaring both sides yields:

$${\cot }^2\theta ={\left(\frac{15}{8}\right)}^2=\frac{225}{64}$$