Simplify fully: \( \sin \left(90^{\circ}-x\right) \cdot \cos \left(180^{\circ}+x\right)+\tan x \cdot \cos x \cdot \sin \left(x-180^{\circ}\right) \) Prove, WITHOUT using a calculator, that \( \frac{\sin 315^{\circ} \cdot \tan 210^{\circ} \cdot \sin 190^{\circ}}{\cos 100^{\circ} \cdot \sin 120^{\circ}}=\frac{-\sqrt{2}}{3} \)

Let's tackle the first expression: 1. Start with the first term: \[ \sin(90^\circ - x) = \cos(x) \] Now consider the second term: \[ \cos(180^\circ + x) = -\cos(x) \] Therefore, we have: \[ \sin(90^{\circ}-x) \cdot \cos(180^{\circ}+x) = \cos(x) \cdot (-\cos(x)) = -\cos^2(x) \] 2. The second part of the expression is: \[ \tan(x) \cdot \cos(x) \cdot \sin(x - 180^\circ) = \tan(x) \cdot \cos(x) \cdot (-\sin(x)) = -\tan(x) \cdot \cos(x) \cdot \sin(x) \] Chaining the definitions, we have: \[ -\tan(x) \cdot \cos(x) \cdot \sin(x) = -\frac{\sin(x)}{\cos(x)} \cdot \cos(x) \cdot \sin(x) = -\sin^2(x) \] 3. Combining both terms: \[ -\cos^2(x) - \sin^2(x) = -(\sin^2(x) + \cos^2(x)) = -1 \] Thus, the final simplified result for the expression is: \[ \sin \left(90^{\circ}-x\right) \cdot \cos \left(180^{\circ}+x\right)+\tan x \cdot \cos x \cdot \sin \left(x-180^{\circ}\right) = -1 \] --- Now, let’s prove the second expression: \[ \frac{\sin 315^{\circ} \cdot \tan 210^{\circ} \cdot \sin 190^{\circ}}{\cos 100^{\circ} \cdot \sin 120^{\circ}} = \frac{-\sqrt{2}}{3} \] 1. Calculate \(\sin 315^\circ\): \[ \sin 315^\circ = \sin(360^\circ - 45^\circ) = -\sin 45^\circ = -\frac{\sqrt{2}}{2} \] 2. Calculate \(\tan 210^\circ\): \[ \tan 210^\circ = \tan(180^\circ + 30^\circ) = \tan 30^\circ = \frac{1}{\sqrt{3}} \text{ (since it's in the 3rd quadrant, it's negative)} = -\frac{1}{\sqrt{3}} \] 3. Calculate \(\sin 190^\circ\): \[ \sin 190^\circ = \sin(180^\circ + 10^\circ) = -\sin 10^\circ \] 4. Calculate \(\cos 100^\circ\): \[ \cos 100^\circ = -\sin 10^\circ \text{ (using co-function identity, } \cos(90^\circ + x) = -\sin x\text{)} \] 5. Calculate \(\sin 120^\circ\): \[ \sin 120^\circ = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2} \] Now substituting all into the left-hand side of our equation gives: \[ \frac{\left(-\frac{\sqrt{2}}{2}\right) \left(-\frac{1}{\sqrt{3}}\right) \left(-\sin 10^\circ\right)}{\left(-\sin 10^\circ\right) \left(\frac{\sqrt{3}}{2}\right)} \] This simplifies as follows: \[ = \frac{\left(\frac{\sqrt{2}}{2\sqrt{3}}\right) (-\sin 10^\circ)}{\left(-\sin 10^\circ\right) \left(\frac{\sqrt{3}}{2}\right)} = \frac{\frac{\sqrt{2}}{2\sqrt{3}}}{\frac{\sqrt{3}}{2}}= \frac{\sqrt{2}}{2\sqrt{3}} \cdot \frac{2}{\sqrt{3}} = \frac{\sqrt{2}}{3} \] However, since we need