(c) Simplify the following: (1) \( \frac{\sin 2 x}{2 \cos ^{2} x} \) (2) \( \frac{\cos 2 \theta}{\sin \theta+\cos \theta} \) (3) \( \frac{1-\cos 2 \mathrm{~A}}{1-\cos ^{2} \mathrm{~A}} \) (4) \( \frac{(\cos \alpha-\sin \alpha)^{2}}{\sin 2 \alpha-1} \) (5) \( \frac{\cos 2 x}{\cos ^{4} x-\sin ^{4} x} \)

1. We start with \[ \frac{\sin 2x}{2 \cos^2 x}. \] Using the double-angle formula \(\sin 2x = 2 \sin x \cos x\), we have \[ \frac{2 \sin x \cos x}{2 \cos^2 x} = \frac{\sin x}{\cos x} = \tan x. \] 2. Consider \[ \frac{\cos 2\theta}{\sin\theta + \cos\theta}. \] Recall the identity \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\), which can be factored as \[ \cos^2 \theta - \sin^2 \theta = (\cos \theta - \sin \theta)(\cos \theta + \sin \theta). \] Thus, \[ \frac{(\cos \theta - \sin \theta)(\cos \theta + \sin \theta)}{\sin \theta + \cos \theta} = \cos \theta - \sin \theta. \] 3. Next, simplify \[ \frac{1-\cos 2A}{1-\cos^2 A}. \] Using the identity \(\cos 2A = 1 - 2\sin^2 A\), we find \[ 1-\cos 2A = 1 - (1-2\sin^2 A) = 2\sin^2 A. \] Also, note that \[ 1-\cos^2 A = \sin^2 A. \] Therefore, \[ \frac{2\sin^2 A}{\sin^2 A} = 2. \] 4. Now simplify \[ \frac{(\cos\alpha - \sin\alpha)^2}{\sin2\alpha - 1}. \] Expand the numerator: \[ (\cos\alpha - \sin\alpha)^2 = \cos^2 \alpha - 2\sin\alpha\cos\alpha + \sin^2 \alpha. \] Since \(\cos^2 \alpha + \sin^2 \alpha = 1\) and \(2\sin\alpha\cos\alpha = \sin2\alpha\), we obtain \[ (\cos\alpha - \sin\alpha)^2 = 1 - \sin2\alpha. \] Thus, \[ \frac{1-\sin2\alpha}{\sin2\alpha-1} = -1. \] 5. Finally, consider \[ \frac{\cos2x}{\cos^4 x-\sin^4 x}. \] Factor the denominator as a difference of squares: \[ \cos^4 x - \sin^4 x = (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x). \] Since \(\cos^2 x + \sin^2 x = 1\), it simplifies to \[ \cos^4 x - \sin^4 x = \cos^2 x - \sin^2 x. \] Recognize that \(\cos2x = \cos^2 x - \sin^2 x\). Thus, \[ \frac{\cos2x}{\cos2x} = 1. \]