Simplify \( \frac{6 \sin \left(180^{\circ}-x\right) \cdot \cos \left(x-360^{\circ}\right)}{\sin ^{2} x-\sin ^{2}\left(90^{\circ}-x\right)} \) to one tri

To simplify \( \frac{6 \sin \left(180^{\circ}-x\right) \cdot \cos \left(x-360^{\circ}\right)}{\sin ^{2} x-\sin ^{2}\left(90^{\circ}-x\right)} \), we start by using trigonometric identities. First, note that: \[ \sin(180^{\circ} - x) = \sin x \] and \[ \cos(x - 360^{\circ}) = \cos x \] Thus, the numerator simplifies to: \[ 6 \sin(180^{\circ} - x) \cdot \cos(x - 360^{\circ}) = 6 \sin x \cdot \cos x \] Now for the denominator, use the identity: \[ \sin^2(90^{\circ} - x) = \cos^2 x \] So, \[ \sin^2 x - \sin^2(90^{\circ} - x) = \sin^2 x - \cos^2 x = \sin^2 x - \cos^2 x = -(\cos^2 x - \sin^2 x) = -\cos(2x) \] Now putting it all together, we have: \[ \frac{6 \sin x \cdot \cos x}{-\cos(2x)} \] Next, recall that: \[ \sin(2x) = 2 \sin x \cos x \] Thus, we have: \[ \frac{6 \sin x \cdot \cos x}{-\cos(2x)} = \frac{3 \cdot 2 \sin x \cos x}{-\cos(2x)} = \frac{-3 \sin(2x)}{\cos(2x)} \] This can be further simplified to: \[ -3 \tan(2x) \] Therefore, the final simplified expression is: \[ \boxed{-3 \tan(2x)} \]