\( \operatorname { Cos } ( 90 ^ { \circ } + x ) \cdot \tan ( 540 ^ { \circ } + x ) \cdot \operatorname { Cos } ( 180 ^ { \circ } - x ) + \operatorname { Sin } ( - 90 ^ { \circ } ) \)

To simplify the expression \( \operatorname{Cos}(90^{\circ} + x) \cdot \tan(540^{\circ} + x) \cdot \operatorname{Cos}(180^{\circ} - x) + \operatorname{Sin}(-90^{\circ}) \), we can use the following identities: 1. **Trigonometric Identities:** - \( \operatorname{Cos}(90^{\circ} + x) = -\operatorname{Sin}(x) \) - \( \operatorname{Tan}(540^{\circ} + x) = \operatorname{Tan}(180^{\circ} + x) = \operatorname{Tan}(x) \) (since the tangent function has a period of \( 180^{\circ} \)) - \( \operatorname{Cos}(180^{\circ} - x) = -\operatorname{Cos}(x) \) - \( \operatorname{Sin}(-90^{\circ}) = -1 \) Now substituting these identities into the expression: \[ -\operatorname{Sin}(x) \cdot \operatorname{Tan}(x) \cdot (-\operatorname{Cos}(x)) - 1 \] This simplifies to: \[ \operatorname{Sin}(x) \cdot \operatorname{Tan}(x) \cdot \operatorname{Cos}(x) - 1 \] Using \( \operatorname{Tan}(x) = \frac{\operatorname{Sin}(x)}{\operatorname{Cos}(x)} \): \[ \operatorname{Sin}(x) \cdot \frac{\operatorname{Sin}(x)}{\operatorname{Cos}(x)} \cdot \operatorname{Cos}(x) - 1 \] This further simplifies to: \[ \operatorname{Sin}^2(x) - 1 \] Therefore the final result is: \[ \operatorname{Sin}^2(x) - 1 \] And since \( \operatorname{Sin}^2(x) - 1 = -\operatorname{Cos}^2(x) \), we can express the final answer as: \[ -\operatorname{Cos}^2(x) \]