\begin{tabular}{l} EXERCISE 9 \\ Simplify the expressions as far as possible. \\ \( \begin{array}{l}1 \quad \frac{\tan \left(180^{\circ}-x\right) \cdot \sin \left(360^{\circ}-x\right) \cdot \cos \left(90^{\circ}-x\right)}{\sin \left(180^{\circ}+x\right) \cdot \cos \left(90^{\circ}+x\right) \cdot \tan \left(180^{\circ}+x\right)} \\ 2 \quad \frac{\sin \left(180^{\circ}+\theta\right) \cdot \cos \left(360^{\circ}-\theta\right) \cdot \tan (-\theta)}{2 \cos \left(90^{\circ}+\theta\right) \cdot \sin \left(90^{\circ}+\theta\right)} \\ \frac{2 \sin \left(90^{\circ}-x\right)+\cos \left(180^{\circ}-x\right)}{\sin \left(90^{\circ}-x\right)-\cos \left(x-180^{\circ}\right)} \\ \cos \left(90^{\circ}-\beta\right)+\cos \left(\beta+90^{\circ}\right)+\cos \left(180^{\circ}-\beta\right)-\cos \left(\beta+180^{\circ}\right)+\cos (-\beta) \\ {\left[\cos \left(90^{\circ}+x\right) \cos (-x) \tan \left(360^{\circ}-x\right)\right]-\left[\cos \left(360^{\circ}-x\right) \sin \left(x-90^{\circ}\right)\right]} \\ {\left[\sin \left(180^{\circ}-x\right)+\sin \left(90^{\circ}-x\right)\right]\left[\cos \left(90^{\circ}-x\right)+\cos (-x)\right]} \\ \\ \frac{\tan \left(180^{\circ}+x\right) \cdot \sin \left(90^{\circ}-x\right)}{\cos \left(90^{\circ}-x\right)}-\frac{\cos \left(180^{\circ}-x\right)}{\sin \left(90^{\circ}+x\right)} \\ \cos \left(90^{\circ}-\theta\right) \cdot \tan \left(180^{\circ}+\theta\right) \cdot \cos \left(360^{\circ}-\theta\right)-\cos (-\theta) \cdot \sin \left(\theta-90^{\circ}\right)\end{array} \) \\ \hline\end{tabular}

1) We have \[ \frac{\tan(180^\circ-x)\cdot\sin(360^\circ-x)\cdot\cos(90^\circ-x)} {\sin(180^\circ+x)\cdot\cos(90^\circ+x)\cdot\tan(180^\circ+x)}. \] Recall: \[ \begin{aligned} \tan(180^\circ-x) &= -\tan x,\\[1mm] \tan(180^\circ+x) &= \tan x,\\[1mm] \sin(360^\circ-x) &= -\sin x,\\[1mm] \sin(180^\circ+x) &= -\sin x,\\[1mm] \cos(90^\circ-x) &= \sin x,\\[1mm] \cos(90^\circ+x) &= -\sin x. \end{aligned} \] Substitute into the expression: \[ \begin{aligned} \text{Numerator} &= (-\tan x)\cdot (-\sin x)\cdot (\sin x) = \tan x\cdot\sin^2 x,\\[1mm] \text{Denominator} &= (-\sin x)\cdot (-\sin x)\cdot (\tan x) = \sin^2 x\cdot\tan x. \end{aligned} \] Thus, \[ \frac{\tan x\cdot\sin^2 x}{\sin^2 x\cdot\tan x}= 1. \] 2) The expression is \[ \frac{\sin(180^\circ+\theta)\cdot\cos(360^\circ-\theta)\cdot\tan(-\theta)} {2\cos(90^\circ+\theta)\cdot\sin(90^\circ+\theta)}. \] Using: \[ \begin{aligned} \sin(180^\circ+\theta) &= -\sin\theta,\\[1mm] \cos(360^\circ-\theta) &= \cos\theta,\\[1mm] \tan(-\theta) &= -\tan\theta,\\[1mm] \cos(90^\circ+\theta) &= -\sin\theta,\\[1mm] \sin(90^\circ+\theta) &= \cos\theta, \end{aligned} \] the numerator becomes: \[ (-\sin\theta)\cdot\cos\theta\cdot(-\tan\theta) =\sin\theta\cos\theta\tan\theta. \] The denominator is: \[ 2\cdot(-\sin\theta)\cdot\cos\theta=-2\sin\theta\cos\theta. \] Since \(\tan\theta=\frac{\sin\theta}{\cos\theta}\), the numerator simplifies to: \[ \sin\theta\cos\theta\cdot\frac{\sin\theta}{\cos\theta}=\sin^2\theta. \] Therefore, the entire expression is: \[ \frac{\sin^2\theta}{-2\sin\theta\cos\theta} =\frac{\sin\theta}{-2\cos\theta} =-\frac{1}{2}\tan\theta. \] 3) We need to simplify \[ \frac{2\sin(90^\circ-x)+\cos(180^\circ-x)} {\sin(90^\circ-x)-\cos(x-180^\circ)}. \] Recall: \[ \begin{aligned} \sin(90^\circ-x)&=\cos x,\\[1mm] \cos(180^\circ-x)&=-\cos x,\\[1mm] \cos(x-180^\circ)&=\cos(180^\circ-x)=-\cos x. \end{aligned} \] Thus the numerator is: \[ 2\cos x+(-\cos x)=\cos x, \] and the denominator is: \[ \cos x-(-\cos x)= 2\cos x. \] So the expression becomes: \[ \frac{\cos x}{2\cos x}=\frac{1}{2}. \] 4) The expression \[ \cos(90^\circ-\beta)+\cos(\beta+90^\circ)+\cos(180^\circ-\beta)-\cos(\beta+180^\circ)+\cos(-\beta) \] simplifies using: \[ \begin{aligned} \cos(90^\circ-\beta)&=\sin\beta,\\[1mm] \cos(\beta+90^\circ)&=-\sin\beta,\\[1mm] \cos(180^\circ-\beta)&=-\cos\beta,\\[1mm] \cos(\beta+180^\circ)&=-\cos\beta,\\[1mm] \cos(-\beta)&=\cos\beta. \end{aligned} \] Then, \[ \sin\beta+(-\sin\beta)=0, \] and \[ \cos(180^\circ-\beta)-\cos(\beta+180^\circ)=(-\cos\beta)-(-\cos\beta)=0. \] Adding the last term, we obtain: \[ 0+ \cos\beta = \cos\beta. \] 5) We have \[ \left[\cos(90^\circ+x)\cos(-x)\tan(360^\circ-x)\right]-\left[\cos(360^\circ-x) \sin(x-90^\circ)\right]. \] Apply: \[ \begin{aligned} \cos(90^\circ+x)&=-\sin x,\\[1mm] \cos(-x)&=\cos x,\\[1mm] \tan(360^\circ-x)&=\tan(-x)=-\tan x,\\[1mm] \cos(360^\circ-x)&=\cos x,\\[1mm] \sin(x-90^\circ)&=-\cos x. \end{aligned} \] Then the first bracket is: \[ (-\sin x)(\cos x)(-\tan x)=\sin x\cos x\,\tan x. \] Since \(\tan x=\frac{\sin x}{\cos x}\), it becomes: \[ \sin x\cos x\,\frac{\sin x}{\cos x}=\sin^2x. \] The second bracket is: \[ \cos x\cdot(-\cos x)=-\cos^2x. \] Thus the whole expression is: \[ \sin^2x -(-\cos^2x) =\sin^2x+\cos^2x=1. \] 6) The expression is \[ \left[\sin(180^\circ-x)+\sin(90^\circ-x)\right] \left[\cos(90^\circ-x)+\cos(-x)\right]. \] Recall: \[ \begin{aligned} \sin(180^\circ