$\displaystyle \tan (\theta )\frac{{dr}}
{{d\theta }} + r = {\sin ^2}(\theta ),\frac{\pi }
{4} < \theta < \frac{\pi }
{2}
$
not sure where to begin...
$\displaystyle \frac{dr}{d\theta}=\frac{\sin^2\theta-r}{\tan\theta}$ $\displaystyle =\sin\theta\cos\theta-r\frac{\cos\theta}{\sin\theta}\Longrightarrow r\theta=\int \sin\theta\cos\theta\,d\theta-r\int \frac{\cos\theta}{\sin\theta}\,d\theta$ ...now end the exercise.
Tonio