(a) Start with the expression \(\left( \frac{1}{\cos x} - \tan x \right) \left( \frac{1}{\sin x} + 1 \right)\).

Use the identity \(\tan x = \frac{\sin x}{\cos x}\) to rewrite \(\tan x\) as \(\frac{\sin x}{\cos x}\).

Substitute to get \(\left( \frac{1}{\cos x} - \frac{\sin x}{\cos x} \right) \left( \frac{1}{\sin x} + 1 \right)\).

Simplify the first term: \(\frac{1 - \sin x}{\cos x}\).

Simplify the second term: \(\frac{1 + \sin x}{\sin x}\).

Multiply the two expressions: \(\frac{(1 - \sin x)(1 + \sin x)}{\cos x \sin x}\).

Use the identity \((1 - \sin x)(1 + \sin x) = 1 - \sin^2 x = \cos^2 x\).

Thus, the expression becomes \(\frac{\cos^2 x}{\cos x \sin x} = \frac{\cos x}{\sin x}\).

Recognize \(\frac{\cos x}{\sin x} = \frac{1}{\tan x}\), proving the identity.

(b) From part (a), we have \(\frac{1}{\tan x} = 2 \tan^2 x\).

Rearrange to get \(\tan^3 x = \frac{1}{2}\).

Take the cube root: \(\tan x = \left( \frac{1}{2} \right)^{1/3}\).

Calculate \(x\) using \(\tan^{-1} \left( \left( \frac{1}{2} \right)^{1/3} \right)\).

Within the range \(0^\circ \leq x \leq 180^\circ\), \(x \approx 38.4^\circ\).