2. (a) Show that $$\frac { 1 - \cos 2 x } { 2 \sin 2 x } \equiv k \tan x \quad x \neq ( 90 n ) ^ { \circ } \quad n \in \mathbb { Z }$$ where \(k\) is a constant to be found.
(b) Hence solve, for \(0 < \theta < 90 ^ { \circ }\) $$\frac { 9 ( 1 - \cos 2 \theta ) } { 2 \sin 2 \theta } = 2 \sec ^ { 2 } \theta$$ giving your answers to one decimal place.
(Solutions based entirely on graphical or numerical methods are not acceptable.)