Simplify the following expression to a single trigonometric ratio: \[ \frac{\sin \left(360^{\circ}-x\right) \cdot \tan (-x)}{\cos \left(180^{\circ}+x\right) \cdot\left(\sin ^{2} A+\cos ^{2} A\right)} \]

To simplify the expression \[ \frac{\sin \left(360^{\circ}-x\right) \cdot \tan (-x)}{\cos \left(180^{\circ}+x\right) \cdot\left(\sin ^{2} A+\cos ^{2} A\right)}, \] we can start by evaluating the trigonometric functions inside the expression. 1. **Evaluate \(\sin(360^{\circ} - x)\)**: \[ \sin(360^{\circ} - x) = -\sin(x) \] 2. **Evaluate \(\tan(-x)\)**: \[ \tan(-x) = -\tan(x) \] 3. **Evaluate \(\cos(180^{\circ} + x)\)**: \[ \cos(180^{\circ} + x) = -\cos(x) \] 4. **Evaluate \(\sin^{2} A + \cos^{2} A\)**: This is a fundamental identity that equals to 1: \[ \sin^{2} A + \cos^{2} A = 1 \] Now substitute these values back into the expression: \[ \frac{\sin(360^{\circ}-x) \cdot \tan(-x)}{\cos(180^{\circ}+x) \cdot\left(\sin^{2} A+\cos^{2} A\right)} = \frac{(-\sin(x)) \cdot (-\tan(x))}{-\cos(x) \cdot 1} \] This simplifies to: \[ \frac{\sin(x) \tan(x)}{\cos(x)} \] Now since \(\tan(x) = \frac{\sin(x)}{\cos(x)}\), we can substitute that into the expression: \[ \frac{\sin(x) \cdot \frac{\sin(x)}{\cos(x)}}{\cos(x)} = \frac{\sin^2(x)}{\cos^2(x)} \] This can be rewritten as: \[ \tan^2(x) \] Thus, the simplified form of the given expression is \[ \boxed{\tan^2(x)}. \]