lify the following without using a \( \frac{\tan \left(-60^{\circ}\right) \cos \left(-156^{\circ}\right) \cos 294^{\circ}}{\sin 492^{\circ}} \) \( \cos ^{2} 375^{\circ}-\cos ^{2}\left(-75^{\circ}\right) \)

Let's simplify both expressions step by step. ### **1. Simplifying \( \frac{\tan(-60^{\circ}) \cos(-156^{\circ}) \cos 294^{\circ}}{\sin 492^{\circ}} \):** 1. **Adjust Angles to Standard Positions:** - \( \tan(-60^{\circ}) = -\tan(60^{\circ}) \) - \( \cos(-156^{\circ}) = \cos(156^{\circ}) \) - \( \sin(492^{\circ}) = \sin(492^{\circ} - 360^{\circ}) = \sin(132^{\circ}) \) 2. **Substitute Known Values:** - \( \tan(60^{\circ}) = \sqrt{3} \) - \( \cos(156^{\circ}) = -\cos(24^{\circ}) \) (since \( 156^{\circ} = 180^{\circ} - 24^{\circ} \)) - \( \cos(294^{\circ}) = \cos(66^{\circ}) \) (since \( 294^{\circ} = 360^{\circ} - 66^{\circ} \)) - \( \sin(132^{\circ}) = \sin(48^{\circ}) \) (since \( 132^{\circ} = 180^{\circ} - 48^{\circ} \)) 3. **Substitute into the Expression:** \[ \frac{-\sqrt{3} \cdot (-\cos(24^{\circ})) \cdot \cos(66^{\circ})}{\sin(48^{\circ})} = \frac{\sqrt{3} \cdot \cos(24^{\circ}) \cdot \cos(66^{\circ})}{\sin(48^{\circ})} \] 4. **Use Product-to-Sum Identity:** \[ \cos(24^{\circ}) \cos(66^{\circ}) = \frac{\cos(90^{\circ}) + \cos(-42^{\circ})}{2} = \frac{0 + \cos(42^{\circ})}{2} = \frac{\cos(42^{\circ})}{2} \] 5. **Simplify Further:** \[ \frac{\sqrt{3} \cdot \frac{\cos(42^{\circ})}{2}}{\sin(48^{\circ})} = \frac{\sqrt{3} \cdot \cos(42^{\circ})}{2 \sin(48^{\circ})} \] Since \( \cos(42^{\circ}) = \sin(48^{\circ}) \): \[ \frac{\sqrt{3} \cdot \sin(48^{\circ})}{2 \sin(48^{\circ})} = \frac{\sqrt{3}}{2} \] ### **2. Simplifying \( \cos^2(375^{\circ}) - \cos^2(-75^{\circ}) \):** 1. **Adjust Angles to Standard Positions:** - \( \cos(375^{\circ}) = \cos(15^{\circ}) \) (since \( 375^{\circ} = 360^{\circ} + 15^{\circ} \)) - \( \cos(-75^{\circ}) = \cos(75^{\circ}) \) 2. **Apply the Difference of Squares Identity:** \[ \cos^2A - \cos^2B = -\sin(A + B) \sin(A - B) \] Let \( A = 15^{\circ} \) and \( B = 75^{\circ} \): \[ \cos^2(15^{\circ}) - \cos^2(75^{\circ}) = -\sin(90^{\circ}) \sin(-60^{\circ}) = -1 \cdot (-\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{2} \] ### **Final Simplified Results:** Both expressions simplify to: \[ \frac{\sqrt{3}}{2} \]