AC = CE ABC is an equilateral triangle BC = AC = CE ∠BCA = 60° ∠BCE = 180° − 60° = 120° BC = CE ∠CBE = ∠CEB = \(\frac{180° − 120°}{2} = 30°\) 180° = \(\pi\) radians 120° = \(\frac{\pi × 120}{180} = \frac{2\pi}{3}\) radians 30° = \(\frac{\pi × 30}{180} = \frac{\pi}{6}\) radians Angles of triangle ACE in radians: \(\frac{2\pi}{3}\), \(\frac{\pi}{6}\), \(\frac{\pi}{6}\)

Q.AC = CE ABC is an equilateral triangle BC = AC = CE ∠BCA = 60° ∠BCE = 180° − 60° = 120° BC = CE ∠CBE = ∠CEB = \(\frac{180° − 120°}{2} = 30°\) 180° = \(\pi\) radians 120° = \(\frac{\pi × 120}{180} = \frac{2\pi}{3}\) radians 30° = \(\frac{\pi × 30}{180} = \frac{\pi}{6}\) radians Angles of triangle ACE in radians: \(\frac{2\pi}{3}\), \(\frac{\pi}{6}\), \(\frac{\pi}{6}\)

Evaluate: \[ \frac{1 - \sin^2 30^\circ}{1 + \sin^2 45^\circ} \times \frac{\cos^2 60^\circ + \cos^2 30^\circ}{\csc^2 90^\circ - \cot^2 90^\circ} \div (\sin 60^\circ \cdot \tan 30^\circ) \] \[ = \frac{1 - \left(\frac{1}{2}\right)^2}{1 + \left(\frac{1}{\sqrt{2}}\right)^2} \times \frac{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}{1^2 - 0^2} \div \left(\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}}\right) \] \[ = \frac{1 - \frac{1}{4}}{1 + \frac{1}{2}} \times \frac{\frac{1}{4} + \frac{3}{4}}{1} \div \frac{1}{2} \] \[ = \frac{3}{4} \div \frac{3}{2} \times 1 \times \frac{2}{1} \] \[ = \frac{3}{4} \cdot \frac{2}{3} \cdot 2 \] \[ = 1 \quad \text{(Answer)} \]