\( \frac { 1 - \cos 2 x } { 2 \sin ^ { 2 } x } + ( \frac { \sin 4 x } { 1 + \cos 4 x } ) ^ { 2 } = \)

Let's simplify the left side of the equation: 1. The first term can be rewritten using the identity \(1 - \cos 2x = 2\sin^2 x\). This gives: \[ \frac{1 - \cos 2x}{2 \sin^2 x} = \frac{2 \sin^2 x}{2 \sin^2 x} = 1. \] 2. Now let's focus on the second term: \[ \frac{\sin 4x}{1 + \cos 4x}. \] We can utilize the identity \(1 + \cos 4x = 2 \cos^2 2x\). Thus, the second term becomes: \[ \frac{\sin 4x}{2 \cos^2 2x}. \] Next, we have: \[ \sin 4x = 2 \sin 2x \cos 2x, \] leading to: \[ \frac{2 \sin 2x \cos 2x}{2 \cos^2 2x} = \frac{\sin 2x}{\cos^2 2x} = \tan 2x \sec 2x. \] Now, we need the square of that: \[ \left( \tan 2x \sec 2x \right)^2 = \tan^2 2x \sec^2 2x. \] 3. This means our original expression simplifies to: \[ 1 + \tan^2 2x \sec^2 2x. \] From the Pythagorean identity, we know that: \[ 1 + \tan^2 \theta = \sec^2 \theta, \] therefore: \[ 1 + \tan^2 2x \sec^2 2x = \sec^2 2x + \tan^2 2x \sec^2 2x = \sec^2 2x (1 + \tan^2 2x) = \sec^2 2x \cdot \sec^2 2x = \sec^4 2x. \] Putting it all together, the final result is: \[ \sec^4 2x. \]