C. $$ \frac{\sin (180-x) an (180-x) \cos (180+x)}{ an (360-x)+a \operatorname{con}(40-x) \sin x} $$

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C. $$ \frac{\sin (180-x) 	an (180-x) \cos (180+x)}{	an (360-x)+a \operatorname{con}(40-x) \sin x} $$

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Let's simplify the given trigonometric expression step by step:

$$
\text{C.} \quad \frac{\sin (180^\circ - x) \cdot \tan (180^\circ - x) \cdot \cos (180^\circ + x)}{\tan (360^\circ - x) + a \cdot \cot(40^\circ - x) \cdot \sin x}
$$

**Step 1: Apply Trigonometric Identities**

1. **Sin and Cosine of Supplementary and Supplementary Angles:**
   - $$\sin(180^\circ - x) = \sin x$$
   - $$\cos(180^\circ + x) = -\cos x$$

2. **Tangent and Cotangent of Angles:**
   - $$\tan(180^\circ - x) = \tan x$$ (since $$\tan$$ has a period of $$180^\circ$$)
   - $$\tan(360^\circ - x) = \tan(-x) = -\tan x$$ (since $$\tan$$ is odd)
   - $$\cot(40^\circ - x) = \frac{1}{\tan(40^\circ - x)}$$

**Step 2: Substitute the Identities into the Expression**

$$
\frac{\sin x \cdot \tan x \cdot (-\cos x)}{-\tan x + a \cdot \cot(40^\circ - x) \cdot \sin x}
$$

Simplify the numerator and the denominator:

- **Numerator:**
  $$
  \sin x \cdot \tan x \cdot (-\cos x) = -\sin x \cdot \left(\frac{\sin x}{\cos x}\right) \cdot \cos x = -\sin^2 x
  $$

- **Denominator:**
  $$
  -\tan x + a \cdot \cot(40^\circ - x) \cdot \sin x = -\tan x + a \cdot \frac{1}{\tan(40^\circ - x)} \cdot \sin x
  $$

**Step 3: Rewrite the Expression**

$$
\frac{-\sin^2 x}{-\tan x + a \cdot \cot(40^\circ - x) \cdot \sin x} = \frac{\sin^2 x}{\tan x - a \cdot \cot(40^\circ - x) \cdot \sin x}
$$

**Step 4: Further Simplification (Optional)**

If further simplification is desired, you can express $$\tan x$$ and $$\cot(40^\circ - x)$$ in terms of sine and cosine:

$$
\tan x = \frac{\sin x}{\cos x} \quad \text{and} \quad \cot(40^\circ - x) = \frac{\cos(40^\circ - x)}{\sin(40^\circ - x)}
$$

Substituting these into the denominator:

$$
\tan x - a \cdot \cot(40^\circ - x) \cdot \sin x = \frac{\sin x}{\cos x} - a \cdot \frac{\cos(40^\circ - x)}{\sin(40^\circ - x)} \cdot \sin x
$$

Factor out $$\sin x$$ from the denominator:

$$
= \sin x \left( \frac{1}{\cos x} - a \cdot \frac{\cos(40^\circ - x)}{\sin(40^\circ - x)} \right)
$$

**Final Simplified Expression:**

$$
\frac{\sin^2 x}{\tan x - a \cdot \cot(40^\circ - x) \cdot \sin x} \quad \text{or} \quad \frac{\sin x}{\frac{1}{\cos x} - a \cdot \frac{\cos(40^\circ - x)}{\sin(40^\circ - x)}}
$$

This is as simplified as the expression can get without additional constraints or information about the variable $$a$$ and the angle $$x$$.
$$ \frac{\sin^2 x}{\tan x - a \cdot \cot(40^\circ - x) \cdot \sin x} $$

C. \( \frac{\sin (180-x) \tan (180-x) \cos (180+x)}{\tan (360-x)+a \operatorname{con}(40-x) \sin x} \)

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