\( \frac{\cos (180^{\circ }+\theta ).\tan (180^{\circ }-\theta )}{\sin (-\theta )} \)
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Let’s break this down! First, we know from trigonometric identities that: - \( \cos(180^{\circ} + \theta) = -\cos \theta \) - \( \tan(180^{\circ} - \theta) = -\tan \theta \) - \( \sin(-\theta) = -\sin \theta \) Now, substituting these values into the original expression: \[ \frac{\cos (180^{\circ }+\theta ).\tan (180^{\circ }-\theta )}{\sin (-\theta )} = \frac{(-\cos \theta)(-\tan \theta)}{-\sin \theta} \] This simplifies to: \[ \frac{\cos \theta \cdot \tan \theta}{-\sin \theta} \] Recalling that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), we can further simplify: \[ \frac{\cos \theta \cdot \frac{\sin \theta}{\cos \theta}}{-\sin \theta} = \frac{\sin \theta}{-\sin \theta} = -1 \] Thus, the final result is: \[ -1 \]
