(a)[4]
Prove that $\dfrac{1 + \sin x}{1 - \sin x} - \dfrac{1 - \sin x}{1 + \sin x} = \dfrac{4 \tan x}{\cos x}$.
(b)[3]
Hence solve the equation $\dfrac{1 + \sin x}{1 - \sin x} - \dfrac{1 - \sin x}{1 + \sin x} = 8 \tan x$ in the interval $0 \leq x \leq \tfrac{1}{2}\pi$.