\( \begin{array}{ll}\text { a) } \frac{\tan \left(360^{\circ}+\theta\right) \cdot \cos (-\theta)}{\cos \left(360^{\circ}-\theta\right)} & \text { (b) } \frac{\sin \left(180^{\circ}-x\right)-\sin (-x)}{\sin \left(360^{\circ}+x\right)} \\ \text { c) } \frac{\sin \left(720^{\circ}-\alpha\right) \cdot \tan (-\alpha)}{\sin \left(720^{\circ}+\alpha\right) \cdot \tan \left(360^{\circ}-\alpha\right)} & \text { (d) } \frac{\cos \left(720^{\circ}+\theta\right)-\cos (-\theta)}{\tan \left(180^{\circ}-\theta\right)} \\ \text { e) } \frac{\sin \left(180^{\circ}-\mathrm{A}\right) \cdot \cos (-\mathrm{A})}{\sin \left(360^{\circ}-\mathrm{A}\right) \cdot \cos \left(\mathrm{A}-180^{\circ}\right)} & \text { (f) } \frac{\tan \left(-x+180^{\circ}\right) \cdot \sin \left(x-360^{\circ}\right)}{\sin \left(360^{\circ}-x\right)} \\ \text { g) } \frac{\sin \left(\theta+180^{\circ}\right) \cdot \tan ^{2}\left(180^{\circ}-\theta\right)}{\tan \left(\theta-180^{\circ}\right) \cdot \sin \left(1080^{\circ}-\theta\right)} & \text { (h) } \frac{\sin ^{2}\left(180^{\circ}+\mathrm{B}\right) \cdot \cos \left(-\mathrm{B}-180^{\circ}\right)}{\sin \left(\mathrm{B}-360^{\circ}\right) \cdot \cos \left(180^{\circ}-\mathrm{B}\right)} \\ \text { i) } \frac{\tan ^{2}\left(x-180^{\circ}\right) \cdot \sin \left(540^{\circ}-x\right)}{\sin \left(180^{\circ}-x\right) \cdot \tan \left(-x-360^{\circ}\right)} & \text { (j) } \frac{\sin \left(\theta-180^{\circ}\right) \cdot \cos \left(540^{\circ}+\theta\right) \cdot \sin (-\theta)}{\cos \left(180^{\circ}-\theta\right) \cdot \sin ^{2}\left(180^{\circ}+\theta\right)}\end{array} \)

\( \textbf{(a)} \) We have \[ \frac{\tan(360^\circ+\theta)\cdot\cos(-\theta)}{\cos(360^\circ-\theta)}. \] Since \[ \tan(360^\circ+\theta)=\tan\theta \quad \text{(tangent has period \(180^\circ\))}, \] \[ \cos(-\theta)=\cos\theta \quad \text{(cosine is even)}, \] and \[ \cos(360^\circ-\theta)=\cos\theta \quad \text{(cosine has period \(360^\circ\))}, \] the expression becomes \[ \frac{\tan\theta\,\cos\theta}{\cos\theta}=\tan\theta. \] \( \textbf{(b)} \) The expression is \[ \frac{\sin(180^\circ-x)-\sin(-x)}{\sin(360^\circ+x)}. \] Note that \[ \sin(180^\circ-x)=\sin x, \] \[ \sin(-x)=-\sin x, \] so the numerator is \[ \sin x-(-\sin x)=2\sin x. \] Also, \[ \sin(360^\circ+x)=\sin x, \] thus \[ \frac{2\sin x}{\sin x}=2. \] \( \textbf{(c)} \) We consider \[ \frac{\sin(720^\circ-\alpha)\cdot\tan(-\alpha)}{\sin(720^\circ+\alpha)\cdot\tan(360^\circ-\alpha)}. \] Since cosine and sine have period \(360^\circ\), \[ \sin(720^\circ-\alpha)=\sin(-\alpha)=-\sin\alpha, \] \[ \sin(720^\circ+\alpha)=\sin\alpha. \] Also, since tangent is odd, \[ \tan(-\alpha)=-\tan\alpha. \] For the remaining term, note that \[ \tan(360^\circ-\alpha)=\tan(-\alpha)=-\tan\alpha. \] Thus the numerator becomes \[ (-\sin\alpha)(-\tan\alpha)=\sin\alpha\,\tan\alpha, \] and the denominator: \[ \sin\alpha\,(-\tan\alpha) = -\sin\alpha\,\tan\alpha. \] Therefore, \[ \frac{\sin\alpha\,\tan\alpha}{-\sin\alpha\,\tan\alpha}=-1. \] \( \textbf{(d)} \) The expression is \[ \frac{\cos(720^\circ+\theta)-\cos(-\theta)}{\tan(180^\circ-\theta)}. \] Since \[ \cos(720^\circ+\theta)=\cos\theta \quad \text{(cosine’s period is \(360^\circ\))}, \] and \[ \cos(-\theta)=\cos\theta, \] the numerator is \[ \cos\theta-\cos\theta=0. \] Thus the whole expression equals \(0\). \( \textbf{(e)} \) We simplify \[ \frac{\sin(180^\circ-A)\cdot\cos(-A)}{\sin(360^\circ-A)\cdot\cos(A-180^\circ)}. \] First, \[ \sin(180^\circ-A)=\sin A \quad \text{and} \quad \cos(-A)=\cos A. \] Next, \[ \sin(360^\circ-A)=\sin(-A)=-\sin A, \] and \[ \cos(A-180^\circ)=\cos(-(180^\circ-A))=\cos(180^\circ-A)=-\cos A. \] Thus the denominator becomes \[ (-\sin A)(-\cos A)=\sin A\,\cos A. \] So the full expression is \[ \frac{\sin A\,\cos A}{\sin A\,\cos A}=1. \] \( \textbf{(f)} \) We have \[ \frac{\tan(-x+180^\circ)\cdot \sin(x-360^\circ)}{\sin(360^\circ-x)}. \] Since \[ \tan(180^\circ-x)= -\tan x, \] we get \[ \tan(-x+180^\circ)=-\tan x. \] Also, \[ \sin(x-360^\circ)=\sin x \quad \text{and} \quad \sin(360^\circ-x)=\sin(-x)=-\sin x. \] Thus, the expression becomes \[ \frac{(-\tan x)\,\sin x}{-\sin x}=\tan x. \] \( \textbf{(g)} \) Consider \[ \frac{\sin(\theta+180^\circ)\cdot \tan^2(180^\circ-\theta)}{\tan(\theta-180^\circ)\cdot\sin(1080^\circ-\theta)}.