Simplify into a single trigonometric ratio. \[ \sqrt{\frac{\frac{1}{2} \sin 2 x}{\tan \left(540^{\circ}+x\right)\left(\frac{1}{\cos ^{2} x}-\tan ^{2} x\right)}} \]

To simplify \( \sqrt{\frac{\frac{1}{2} \sin 2 x}{\tan \left(540^{\circ}+x\right)\left(\frac{1}{\cos ^{2} x}-\tan ^{2} x\right)}} \), we start off by addressing each component step-by-step. 1. **Simplifying \(\tan(540^\circ + x)\)**: \[ \tan(540^\circ + x) = \tan(180^\circ + 360^\circ + x) = \tan(x) \] (since \( \tan(\theta + 180^\circ) = \tan(\theta) \)). 2. **Simplifying \(\frac{1}{\cos^2 x} - \tan^2 x\)**: Recall that \(\tan x = \frac{\sin x}{\cos x}\): \[ \tan^2 x = \frac{\sin^2 x}{\cos^2 x} \] Thus, \[ \frac{1}{\cos^2 x} - \tan^2 x = \frac{1}{\cos^2 x} - \frac{\sin^2 x}{\cos^2 x} = \frac{1 - \sin^2 x}{\cos^2 x} = \frac{\cos^2 x}{\cos^2 x} = 1 \] 3. **Now substitute these results back into the original expression**: \[ \sqrt{\frac{\frac{1}{2} \sin 2x}{\tan x \cdot 1}} = \sqrt{\frac{\frac{1}{2} \sin 2x}{\tan x}} \] 4. **Recalling that \(\sin 2x = 2 \sin x \cos x\)** and \(\tan x = \frac{\sin x}{\cos x}\): \[ \sqrt{\frac{\frac{1}{2} \cdot 2 \sin x \cos x}{\frac{\sin x}{\cos x}}} = \sqrt{\frac{\sin x \cos x}{\frac{\sin x}{\cos x}}} \] 5. **This simplifies further** (as long as \(\sin x \neq 0\)): \[ \sqrt{\sin x \cos^2 x} = \sqrt{\sin x} \cdot \cos x \] In conclusion, we obtain the simplified trigonometric ratio: \[ \sqrt{\sin x} \cdot \cos x \]