1. Problem: Simplify each expression involving trigonometric functions of angle $\alpha$. 2. a) $1 - \cos^2 \alpha - \sin^2 \alpha = 1 - (\cos^2 \alpha + \sin^2 \alpha) = 1 - 1 = 0$ 3. b) $\tan \alpha \cdot \cos \alpha - \sin \alpha = \frac{\sin \alpha}{\cos \alpha} \cdot \cos \alpha - \sin \alpha = \sin \alpha - \sin \alpha = 0$ 4. c) $\frac{1 - \sin^2 \alpha}{\cos \alpha} = \frac{\cos^2 \alpha}{\cos \alpha} = \cos \alpha$ 5. d) $\frac{1 - \cos^2 \alpha}{\sin \alpha \cdot \cos \alpha} = \frac{\sin^2 \alpha}{\sin \alpha \cdot \cos \alpha} = \frac{\sin \alpha}{\cos \alpha} = \tan \alpha$ 6. e) Same as b), so result is $0$ 7. f) $\cos^2 \alpha \cdot (1 + \tan^2 \alpha) = \cos^2 \alpha \cdot \frac{1}{\cos^2 \alpha} = 1$ 8. g) $\frac{\sin \alpha}{1 - \cos \alpha} + \frac{\sin \alpha}{1 + \cos \alpha} = \sin \alpha \left( \frac{1}{1 - \cos \alpha} + \frac{1}{1 + \cos \alpha} \right) = \sin \alpha \cdot \frac{(1 + \cos \alpha) + (1 - \cos \alpha)}{(1 - \cos \alpha)(1 + \cos \alpha)} = \sin \alpha \cdot \frac{2}{1 - \cos^2 \alpha} = \sin \alpha \cdot \frac{2}{\sin^2 \alpha} = \frac{2}{\sin \alpha}$ 9. h) $\sin(1440^\circ + \alpha) + \sin(-\alpha)$. Since $1440^\circ = 4 \times 360^\circ$, $\sin(1440^\circ + \alpha) = \sin \alpha$. Also, $\sin(-\alpha) = -\sin \alpha$. Sum is $\sin \alpha - \sin \alpha = 0$ 10. i) $6 \sin^2(-\alpha) + 6 \cos^2(-\alpha) - 5 = 6 \sin^2 \alpha + 6 \cos^2 \alpha - 5 = 6(\sin^2 \alpha + \cos^2 \alpha) - 5 = 6 \times 1 - 5 = 1$ 11. j) $\sin^3 \alpha + \sin \alpha \cdot \cos^2 \alpha + \sin(-\alpha) = \sin^3 \alpha + \sin \alpha (1 - \sin^2 \alpha) - \sin \alpha = \sin^3 \alpha + \sin \alpha - \sin^3 \alpha - \sin \alpha = 0$ 12. k) $(\sin \alpha + \cos \alpha)^2 + (\sin \alpha - \cos \alpha)^2 = (\sin^2 \alpha + 2 \sin \alpha \cos \alpha + \cos^2 \alpha) + (\sin^2 \alpha - 2 \sin \alpha \cos \alpha + \cos^2 \alpha) = 2(\sin^2 \alpha + \cos^2 \alpha) = 2 \times 1 = 2$ 13. l) $(1 - \sin^2 \alpha) \cdot \tan^2 \alpha + \cos^2 \alpha = \cos^2 \alpha \cdot \frac{\sin^2 \alpha}{\cos^2 \alpha} + \cos^2 \alpha = \sin^2 \alpha + \cos^2 \alpha = 1$ 14. m) $\frac{\sin \alpha \cdot \cos(-\alpha)}{1 - \sin^2 \alpha} - \tan \alpha = \frac{\sin \alpha \cdot \cos \alpha}{\cos^2 \alpha} - \frac{\sin \alpha}{\cos \alpha} = \frac{\sin \alpha}{\cos \alpha} - \frac{\sin \alpha}{\cos \alpha} = 0$ 15. n) $(1 - \sin \alpha)(1 + \sin \alpha) = 1 - \sin^2 \alpha = \cos^2 \alpha$ 16. o) $\cos^2 \alpha + \sin^2 \alpha - 6 = 1 - 6 = -5$ 17. p) $\frac{1}{\sin^2 \alpha} - \sin^2 \alpha \cdot \frac{1}{\cos \alpha} = \csc^2 \alpha - \frac{\sin^2 \alpha}{\cos \alpha}$ (cannot simplify further without specific values) 18. q) $\frac{(\sin \alpha - \cos \alpha)^2 - 1}{1 - \sin^2 \alpha} = \frac{(\sin^2 \alpha - 2 \sin \alpha \cos \alpha + \cos^2 \alpha) - 1}{\cos^2 \alpha} = \frac{(1 - 2 \sin \alpha \cos \alpha) - 1}{\cos^2 \alpha} = \frac{-2 \sin \alpha \cos \alpha}{\cos^2 \alpha} = -2 \tan \alpha$ Final answers: a) 0 b) 0 c) $\cos \alpha$ d) $\tan \alpha$ e) 0 f) 1 g) $\frac{2}{\sin \alpha}$ h) 0 i) 1 j) 0 k) 2 l) 1 m) 0 n) $\cos^2 \alpha$ o) -5 p) $\csc^2 \alpha - \frac{\sin^2 \alpha}{\cos \alpha}$ q) $-2 \tan \alpha$