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

To simplify the expression \( C = \frac{\sin (180-x) \tan (180-x) \cos (180+x)}{\tan (360-x)+a \operatorname{con}(40-x) \sin x} \), we can use trigonometric identities. 1. **Simplify the Numerator**: - \(\sin(180-x) = \sin x\) - \(\tan(180-x) = -\tan x\) - \(\cos(180+x) = -\cos x\) Therefore, the numerator simplifies to: \[ \sin(180-x) \tan(180-x) \cos(180+x) = \sin x (-\tan x)(-\cos x) = \sin x \tan x \cos x \] Recall that \(\tan x = \frac{\sin x}{\cos x}\): \[ \sin x \tan x \cos x = \sin x \cdot \frac{\sin x}{\cos x} \cdot \cos x = \sin^2 x \] 2. **Simplify the Denominator**: - \(\tan(360-x) = -\tan x\), since \(360-x = 360^{\circ} - x\) - The term stays as \(\tan(360-x) = -\tan x\). Thus, the denominator becomes: \[ \tan(360-x)+a \operatorname{con}(40-x) \sin x = -\tan x + a \operatorname{con}(40-x) \sin x \] 3. **Final Expression**: Putting it all together, the expression for \( C \) simplifies to: \[ C = \frac{\sin^2 x}{-\tan x + a \operatorname{con}(40-x) \sin x} \] This is the simplified form of the given expression.