12. Simplify the following: \( \begin{array}{ll}\text { (a) } \frac{\sin \left(180^{\circ}+2 x\right)}{\cos \left(180^{\circ}+x\right)} & \text { (b) } \frac{\tan 45^{\circ}-\sin ^{2}(-x)}{\sin ^{2}\left(90^{\circ}+x\right)} \\ \text { (c) } \frac{\sin 4 x}{1-2 \cos ^{2} 2 x} & \text { (d) } \frac{\sin \left(-180^{\circ}-\alpha\right) \tan (-\alpha) \cos \left(360^{\circ}-\alpha\right)}{\cos ^{2}\left(90^{\circ}+\alpha\right)+\sin ^{2}\left(180^{\circ}+\alpha\right)}\end{array} \)

**(a)** We start with \[ \frac{\sin\left(180^\circ+2x\right)}{\cos\left(180^\circ+x\right)}. \] Recall that \[ \sin(180^\circ+\theta) = -\sin\theta \quad \text{and} \quad \cos(180^\circ+\theta) = -\cos\theta. \] Thus, \[ \sin\left(180^\circ+2x\right) = -\sin2x \quad \text{and} \quad \cos\left(180^\circ+x\right) = -\cos x. \] Substitute these into the expression: \[ \frac{-\sin2x}{-\cos x} = \frac{\sin2x}{\cos x}. \] Now, use the double-angle formula for sine: \[ \sin2x = 2\sin x\,\cos x. \] So, \[ \frac{2\sin x\,\cos x}{\cos x} = 2\sin x. \] --- **(b)** We have \[ \frac{\tan45^\circ - \sin^2(-x)}{\sin^2(90^\circ+x)}. \] Since \[ \tan45^\circ = 1, \] and using that sine is an odd function, \[ \sin(-x) = -\sin x \quad \Rightarrow \quad \sin^2(-x) = \sin^2x. \] Thus, the numerator becomes: \[ 1 - \sin^2x. \] Next, recall the co-function identity: \[ \sin(90^\circ+x) = \cos x, \] so \[ \sin^2(90^\circ+x) = \cos^2x. \] The expression now is: \[ \frac{1-\sin^2x}{\cos^2x}. \] Use the Pythagorean identity: \[ 1-\sin^2x=\cos^2x. \] Thus, the expression simplifies to: \[ \frac{\cos^2x}{\cos^2x} = 1. \] --- **(c)** We need to simplify \[ \frac{\sin4x}{1-2\cos^2 2x}. \] Notice that the cosine double-angle formula gives: \[ \cos4x = 2\cos^2 2x-1. \] Rearrange this to express \(1-2\cos^2 2x\): \[ 1-2\cos^2 2x = -(2\cos^2 2x-1)=-\cos4x. \] Substitute back into the original expression: \[ \frac{\sin4x}{-\cos4x} = -\frac{\sin4x}{\cos4x} = -\tan4x. \] --- **(d)** We simplify \[ \frac{\sin(-180^\circ-\alpha)\,\tan(-\alpha)\,\cos(360^\circ-\alpha)}{\cos^2(90^\circ+\alpha)+\sin^2(180^\circ+\alpha)}. \] *Numerator:* - Use the odd property of sine: \[ \sin(-180^\circ-\alpha) = -\sin(180^\circ+\alpha). \] And note that \[ \sin(180^\circ+\alpha) = -\sin\alpha, \] so \[ \sin(-180^\circ-\alpha) = -(-\sin\alpha)=\sin\alpha. \] - For tangent, since it is odd: \[ \tan(-\alpha) = -\tan\alpha. \] - For cosine, use the periodic property: \[ \cos(360^\circ-\alpha)=\cos\alpha. \] Thus, the numerator becomes: \[ \sin\alpha\cdot(-\tan\alpha)\cdot\cos\alpha = -\sin\alpha\,\tan\alpha\,\cos\alpha. \] Since \[ \tan\alpha=\frac{\sin\alpha}{\cos\alpha}, \] the numerator simplifies to: \[ -\sin\alpha\cdot\frac{\sin\alpha}{\cos\alpha}\cdot\cos\alpha = -\sin^2\alpha. \] *Denominator:* - We know: \[ \cos(90^\circ+\alpha) = -\sin\alpha \quad \Rightarrow \quad \cos^2(90^\circ+\alpha)=\sin^2\alpha. \] - And \[ \sin(180^\circ+\alpha) = -\sin\alpha \quad \Rightarrow \quad \sin^2(180^\circ+\alpha)=\sin^2\alpha. \] Thus, the denominator becomes: \[ \sin^2\alpha+\sin^2\alpha = 2\sin^2\alpha. \] Combining numerator and denominator yields: \[ \frac{-\sin^2\alpha}{2\sin^2\alpha}=-\frac{1}{2}. \] --- **Final Answers:** (a) \(2\sin x\) (b) \(1\) (c) \(-\tan4x\) (d) \(-\frac{1}{2}\)