Question 4 - A-Level Maths

Prove the identity $\frac { \sin ^ { 3 } \theta } { \sin \theta - 1 } - \frac { \sin ^ { 2 } \theta } { 1 + \sin \theta } \equiv - \tan ^ { 2 } \theta \left( 1 + \sin ^ { 2 } \theta \right)$.
Hence solve the equation $$\frac { \sin ^ { 3 } \theta } { \sin \theta - 1 } - \frac { \sin ^ { 2 } \theta } { 1 + \sin \theta } = \tan ^ { 2 } \theta \left( 1 - \sin ^ { 2 } \theta \right)$$ for $0 < \theta < 2 \pi$.