(a) Prove the identity \(\frac{1 + \sin x}{1 - \sin x} - \frac{1 - \sin x}{1 + \sin x} \equiv \frac{4 \tan x}{\cos x}\).

(b) Hence solve the equation \(\frac{1 + \sin x}{1 - \sin x} - \frac{1 - \sin x}{1 + \sin x} = 8 \tan x\) for \(0 \leq x \leq \frac{1}{2} \pi\).