Question 7 - A-Level Maths

Show that \(\frac { \sin \theta } { 1 - \sin \theta } - \frac { \sin \theta } { 1 + \sin \theta } \equiv 2 \tan ^ { 2 } \theta\).
Hence solve the equation \(\frac { \sin \theta } { 1 - \sin \theta } - \frac { \sin \theta } { 1 + \sin \theta } = 8\), for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).