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

Step 1 ：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. The given expression is $\frac{1+{\cot }^2\theta }{{\csc }^2\theta -1}$.

Step 2 ：Convert the cotangent and cosecant functions into their equivalent sine and cosine functions. Cotangent is the reciprocal of tangent, which is sine over cosine, and cosecant is the reciprocal of sine. So, we can rewrite the expression as follows: $\frac{1+{\left(\frac{\cos \theta }{\sin \theta }\right)}^2}{{\left(\frac{1}{\sin \theta }\right)}^2-1}$.

Step 3 ：Simplify the expression by multiplying the numerator and denominator by ${\sin }^2\theta$ to eliminate the fractions within the fractions. This will give us: $\frac{{\sin }^2\theta +{\cos }^2\theta }{1-{\sin }^2\theta }$.

Step 4 ：Use the Pythagorean identity ${\sin }^2\theta +{\cos }^2\theta =1$ to further simplify the expression to: $\frac{1}{1-{\sin }^2\theta }$.

Step 5 ：The simplified expression in terms of sine and cosine, with no quotients appearing in the final expression and all functions are of $\theta$ only is: $\boxed{{\displaystyle \frac{1}{1-{\sin }^2\theta }}}$.