1. The problem asks us to find the derivative of the function $$r = (\csc \theta + \cot \theta)^{-1}$$ with respect to $$\theta$$. 2. Recall the derivative rules: the derivative of $$u^{-1}$$ is $$-u^{-2} \frac{du}{d\theta}$$. 3. Let $$u = \csc \theta + \cot \theta$$. Then we need $$\frac{du}{d\theta}$$. 4. Derivative of $$\csc \theta$$ is $$-\csc \theta \cot \theta$$. 5. Derivative of $$\cot \theta$$ is $$-\csc^{2} \theta$$. 6. So, $$\frac{du}{d\theta} = -\csc \theta \cot \theta - \csc^{2} \theta$$. 7. Thus, by the chain rule, $$\frac{dr}{d\theta} = -u^{-2} \frac{du}{d\theta} = - (\csc \theta + \cot \theta)^{-2} \left(-\csc \theta \cot \theta - \csc^{2} \theta \right)$$. 8. Simplify the negative signs: $$\frac{dr}{d\theta} = (\csc \theta + \cot \theta)^{-2} (\csc \theta \cot \theta + \csc^{2} \theta)$$. 9. We can factor $$\csc \theta$$ inside the parenthesis: $$\frac{dr}{d\theta} = (\csc \theta + \cot \theta)^{-2} \csc \theta (\cot \theta + \csc \theta)$$. 10. Recognizing $$\cot \theta + \csc \theta$$ is the same as $$u$$, we have: $$\frac{dr}{d\theta} = (\csc \theta + \cot \theta)^{-2} \csc \theta (\csc \theta + \cot \theta) = \csc \theta (\csc \theta + \cot \theta)^{-1}.$$