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This problem involves simplifying the trigonometric expression:  

\[
\frac{\cos 105^\circ - \sin 15^\circ}{\cos 165^\circ - \cos 75^\circ}.
\]

### Solution:

1. **Rewrite trigonometric angles using identities:**

   \(\cos 105^\circ\) and \(\sin 15^\circ\) can be related using complementary angles:  
   \[
   \cos 105^\circ = \cos (90^\circ + 15^\circ) = -\sin 15^\circ.
   \]
   Therefore:
   \[
   \cos 105^\circ - \sin 15^\circ = -\sin 15^\circ - \sin 15^\circ = -2 \sin 15^\circ.
   \]

   For the denominator, use the **difference of cosines identity**:
   \[
   \cos A - \cos B = -2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right).
   \]
   Here, \(A = 165^\circ\) and \(B = 75^\circ\):
   \[
   \cos 165^\circ - \cos 75^\circ = -2 \sin \left(\frac{165^\circ + 75^\circ}{2}\right) \sin \left(\frac{165^\circ - 75^\circ}{2}\right).
   \]
   Simplify:
   \[
   \cos 165^\circ - \cos 75^\circ = -2 \sin (120^\circ) \sin (45^\circ).
   \]

2. **Simplify the trigonometric values:**

   Recall standard trigonometric values:
   \[
   \sin 15^\circ, \sin 120^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2}, \text{ and } \sin 45^\circ = \frac{\sqrt{2}}{2}.
   \]

   Substituting into the denominator:
   \[
   \cos 165^\circ - \cos 75^\circ = -2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} = -\frac{\sqrt{6}}{2}.
   \]

   For the numerator:
   \[
   \cos 105^\circ - \sin 15^\circ = -2 \sin 15^\circ.
   \]

3. **Combine numerator and denominator:**

   The expression becomes:
   \[
   \frac{\cos 105^\circ - \sin 15^\circ}{\cos 165^\circ - \cos 75^\circ} = \frac{-2 \sin 15^\circ}{-\frac{\sqrt{6}}{2}}.
   \]

   Simplify:
   \[
   \frac{-2 \sin 15^\circ}{-\frac{\sqrt{6}}{2}} = \frac{4 \sin 15^\circ}{\sqrt{6}}.
   \]

4. **Simplify further if necessary:**

   The exact value of \(\sin 15^\circ\) is:
   \[
   \sin 15^\circ = \sin (45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ.
   \]
   Using known values:
   \[
   \sin 15^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}.
   \]

   Substituting:
   \[
   \frac{4 \sin 15^\circ}{\sqrt{6}} = \frac{4 \cdot \frac{\sqrt{6} - \sqrt{2}}{4}}{\sqrt{6}} = \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6}}.
   \]

5. **Final result:**
   \[
   \frac{\cos 105^\circ - \sin 15^\circ}{\cos 165^\circ - \cos 75^\circ} = 1 - \frac{\sqrt{2}}{\sqrt{6}}.
   \]

Would you like more explanation or a detailed step-by-step breakdown of a specific part?  

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### Related Questions:
1. How can complementary angle identities simplify trigonometric expressions?
2. What is the geometric interpretation of \(\cos(105^\circ)\) and \(\sin(15^\circ)\)?
3. How does the cosine difference identity apply to real-world problems?
4. Why is it important to memorize exact trigonometric values for common angles?
5. Can \(\sin(15^\circ)\) be calculated directly without using identities?

### Tip:
Remember key trigonometric identities like sum and difference formulas to simplify problems involving uncommon angles.

Find the value of the expression \(\frac{\cos 105^\circ - \sin 15^\circ}{\cos 165^\circ - \cos 75^\circ}\).

This solution simplifies the trigonometric expression \(\frac{\cos 105^\circ - \sin 15^\circ}{\cos 165^\circ - \cos 75^\circ}\) by applying complementary angle identities and the difference of cosines formula. Explore the detailed step-by-step process, including exact values for \(\sin(15^\circ)\), \(\sin(120^\circ)\), and \(\sin(45^\circ)\). Ideal for Grades 10-12.

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