Answer
$$r = \frac{1}{3}{\sin ^2}\theta + C\csc \theta $$
Work Step by Step
$$\eqalign{
& \tan \theta \frac{{dr}}{{d\theta }} + r = {\sin ^2}\theta ,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 < \theta < \pi /2 \cr
& {\text{Divide each term of the differential equation by }}\tan \theta \cr
& \frac{{\tan \theta }}{{\tan \theta }}\frac{{dr}}{{d\theta }} + \frac{r}{{\tan \theta }} = \frac{{{{\sin }^2}\theta }}{{\tan \theta }} \cr
& \frac{{dr}}{{d\theta }} + \frac{{\cos \theta }}{{\sin \theta }}r = \sin \theta \cos \theta \,\,\,\,\,\,\left( 1 \right) \cr
& {\text{The equation is written in the form }}\frac{{dr}}{{d\theta }} + P\left( \theta \right)r = Q\left( \theta \right) \cr
& P\left( \theta \right) = \frac{{\cos \theta }}{{\sin \theta }},\,\,\,Q\left( \theta \right) = \sin \theta \cos \theta \,\,\,\,\,\,\, \cr
& {\text{The integrating factor is }}\cr
& v\left( \theta \right) = {e^{\int {\frac{{\cos \theta }}{{\sin \theta }}} d\theta }} = {e^{\ln \left| {\sin \theta } \right|}} = \sin \theta \cr
& {\text{Multiply equation 1 by the integrating factor}} \cr
& \sin \theta \frac{{dr}}{{d\theta }} + \sin \theta \frac{{\cos \theta }}{{\sin \theta }}r = \sin \theta \sin \theta \cos \theta \,\,\,\,\,\, \cr
& \sin \theta \frac{{dr}}{{d\theta }} + \cos \theta r = {\sin ^2}\theta \,\cos \theta \,\,\,\,\, \cr
& \cr
& {\text{Write the left - hand in the form }}\frac{d}{{d\theta }}\left[ {r\sin \theta } \right] \cr
& \frac{d}{{d\theta }}\left[ {r\sin \theta } \right] = {\sin ^2}\theta \cos \theta \,\,\,\,\, \cr
& \cr
& {\text{Integrate both sides of the differential equation}} \cr
& r\sin \theta = \int {{{\sin }^2}\theta \cos \theta \,} d\theta \cr
& r\sin \theta = \frac{1}{3}{\sin ^3}\theta + C \cr
& {\text{Solve for }}r \cr
& r = \frac{1}{3}\frac{{{{\sin }^3}\theta }}{{\sin \theta }} + \frac{C}{{\sin \theta }} \cr
& r = \frac{1}{3}{\sin ^2}\theta + C\csc \theta \cr} $$
