č) \( \frac{1-\cos 2 x}{2 \sin ^{2} x}+\left(\frac{\sin 4 x}{1+\cos 4 x}\right)^{2} \)

To simplify the expression \( \frac{1-\cos 2x}{2 \sin^2 x} + \left(\frac{\sin 4x}{1+\cos 4x}\right)^2 \), we will first work on each part separately. 1. **First Term**: The first term involves \( \frac{1-\cos 2x}{2 \sin^2 x} \). We can use the identity \( 1 - \cos 2x = 2 \sin^2 x \): \[ \frac{1 - \cos 2x}{2 \sin^2 x} = \frac{2 \sin^2 x}{2 \sin^2 x} = 1 \] 2. **Second Term**: The second term is \( \left(\frac{\sin 4x}{1+\cos 4x}\right)^2 \). We will simplify \( \frac{\sin 4x}{1+\cos 4x} \) using the half-angle identity: Since \( 1+\cos 4x = 2\cos^2 2x \) and \( \sin 4x = 2\sin 2x \cos 2x \), we can rewrite it as: \[ \frac{\sin 4x}{1+\cos 4x} = \frac{2 \sin 2x \cos 2x}{2 \cos^2 2x} = \frac{\sin 2x}{\cos^2 2x} = \tan 2x \sec 2x \] Squaring this gives: \[ \left(\frac{\sin 4x}{1+\cos 4x}\right)^2 = \tan^2 2x \sec^2 2x = \tan^2 2x (1 + \tan^2 2x) = \tan^2 2x + \tan^4 2x \] 3. **Combining the Results**: Now combine the results: The whole expression becomes: \[ 1 + \tan^2 2x + \tan^4 2x \] To summarize, the original expression simplifies to: \[ 1 + \tan^2 2x + \tan^4 2x \]