Answer
$$\frac{{dr}}{{d\theta }} = \frac{{\csc \theta }}{{\cot \theta + \csc \theta }}$$
Work Step by Step
$$\eqalign{
& r = {\left( {\csc \theta + \cot \theta } \right)^{ - 1}} \cr
& {\text{differentiate with respect to }}\theta \cr
& \frac{{dr}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {{{\left( {\csc \theta + \cot \theta } \right)}^{ - 1}}} \right] \cr
& \cr
& {\text{using the chain rule }}\frac{d}{{d\theta }}\left[ {{u^n}} \right] = n{u^{n - 1}}\frac{{du}}{{d\theta }} \cr
& \frac{{dr}}{{d\theta }} = - {\left( {\csc \theta + \cot \theta } \right)^{ - 2}}\frac{d}{{d\theta }}\left[ {\csc \theta + \cot \theta } \right] \cr
& \cr
& {\text{solving the derivatives}} \cr
& \frac{{dr}}{{d\theta }} = - {\left( {\csc \theta + \cot \theta } \right)^{ - 2}}\left( { - \csc \theta \cot \theta - {{\csc }^2}\theta } \right) \cr
& \cr
& {\text{simplifying, we get:}} \cr
& \frac{{dr}}{{d\theta }} = {\left( {\csc \theta + \cot \theta } \right)^{ - 2}}\left( {\csc \theta \cot \theta + {{\csc }^2}\theta } \right) \cr
& \frac{{dr}}{{d\theta }} = \frac{{\csc \theta \cot \theta + {{\csc }^2}\theta }}{{{{\left( {\csc \theta + \cot \theta } \right)}^2}}} \cr
& \frac{{dr}}{{d\theta }} = \frac{{\csc \theta \left( {\cot \theta + \csc \theta } \right)}}{{{{\left( {\csc \theta + \cot \theta } \right)}^2}}} \cr
& \frac{{dr}}{{d\theta }} = \frac{{\csc \theta }}{{\cot \theta + \csc \theta }} \cr} $$
