Write each expression in terms of sine and cosine, and then simplify so that no quotients appear in the final expression and all functions are of $\theta$ only.
$${\sin }^2(-\theta )+{\cot }^2(-\theta )+{\cos }^2(-\theta )$$
$${\sin }^2(-\theta )+{\cot }^2(-\theta )+{\cos }^2(-\theta )=$$

Step 1 ：First, we write each function in terms of sine and cosine. We have ${\sin }^2(-\theta )={\sin }^2\theta$, ${\cos }^2(-\theta )={\cos }^2\theta$, and ${\cot }^2(-\theta )={\cot }^2\theta$.

Step 2 ：Next, we express ${\cot }^2\theta$ in terms of sine and cosine. We know that $\cot \theta =\frac{\cos \theta }{\sin \theta }$, so ${\cot }^2\theta =\frac{{\cos }^2\theta }{{\sin }^2\theta }$.

Step 3 ：Substituting these into the original expression, we get ${\sin }^2\theta +\frac{{\cos }^2\theta }{{\sin }^2\theta }+{\cos }^2\theta$.

Step 4 ：To simplify this expression, we multiply the second term by ${\sin }^2\theta$ to get ${\sin }^2\theta +{\cos }^2\theta +{\cos }^2\theta {\sin }^2\theta$.

Step 5 ：Finally, we use the Pythagorean identity ${\sin }^2\theta +{\cos }^2\theta =1$ to simplify the expression to $1+{\cos }^2\theta {\sin }^2\theta$.

Step 6 ：Since ${\sin }^2\theta$ and ${\cos }^2\theta$ are both less than or equal to 1, the final expression is always less than or equal to 2. Therefore, the simplified expression is $\boxed{{\displaystyle 1+{\cos }^2\theta {\sin }^2\theta }}$.