(a) To prove the identity:

Start with \(\frac{1 + \sin x}{1 - \sin x} - \frac{1 - \sin x}{1 + \sin x}\).

Use a common denominator: \((1 - \sin x)(1 + \sin x) = 1 - \sin^2 x = \cos^2 x\).

Numerator becomes \((1 + \sin x)^2 - (1 - \sin x)^2 = 1 + 2\sin x + \sin^2 x - (1 - 2\sin x + \sin^2 x) = 4\sin x\).

Thus, \(\frac{4\sin x}{\cos^2 x} = \frac{4\sin x}{\cos x \cdot \cos x} = \frac{4\tan x}{\cos x}\).

Identity is proven.

(b) Solve \(\frac{4\tan x}{\cos x} = 8\tan x\).

Divide both sides by \(\tan x\) (assuming \(\tan x \neq 0\)): \(\frac{4}{\cos x} = 8\).

\(\cos x = \frac{1}{2}\).

\(x = \frac{\pi}{3}\) within the given range.

Also, consider \(\tan x = 0\) which gives \(x = 0\).

Thus, solutions are \(x = 0\) or \(x = \frac{\pi}{3}\).