1. Problem 21: Simplify $$\cos(87\pi + \alpha) \cdot \cos\left(\frac{57\pi}{2} - \alpha\right) \cdot \tan(2025\pi + \alpha)$$. - Use periodicity: $$\cos(87\pi + \alpha) = \cos(\alpha)$$ because $$\cos(\theta + 2k\pi) = \cos \theta$$ and $$87\pi = 43\cdot 2\pi + \pi$$, so $$\cos(87\pi + \alpha) = \cos(\pi + \alpha) = -\cos \alpha$$. - Similarly, $$\cos\left(\frac{57\pi}{2} - \alpha\right) = \cos\left(28\pi + \frac{\pi}{2} - \alpha\right) = \cos\left(\frac{\pi}{2} - \alpha\right) = \sin \alpha$$. - For $$\tan(2025\pi + \alpha)$$, since $$2025\pi = 1012\cdot 2\pi + \pi$$, $$\tan(2025\pi + \alpha) = \tan(\pi + \alpha) = \tan \alpha$$. - So the expression becomes $$(-\cos \alpha) \cdot \sin \alpha \cdot \tan \alpha = -\cos \alpha \cdot \sin \alpha \cdot \frac{\sin \alpha}{\cos \alpha} = -\sin^2 \alpha$$. Answer: none of the options exactly match $$-\sin^2 \alpha$$, but closest is D (sin³ α) if sign ignored. Since none match, the simplified form is $$-\sin^2 \alpha$$. 2. Problem 22: Simplify $$\frac{(\cos \alpha + 1)(1 - \cos \alpha)}{\tan^2 \alpha}$$. - Numerator: $$(\cos \alpha + 1)(1 - \cos \alpha) = 1 - \cos^2 \alpha = \sin^2 \alpha$$. - Denominator: $$\tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha}$$. - So expression is $$\frac{\sin^2 \alpha}{\frac{\sin^2 \alpha}{\cos^2 \alpha}} = \cos^2 \alpha$$. Answer: A (cos² α). 3. Problem 23: Given $$\sin \alpha = m$$ and $$0 < \alpha < \frac{\pi}{2}$$, find $$\sin \alpha \cdot \cos^3 \alpha \cdot \tan \alpha$$. - Express $$\cos \alpha = \sqrt{1 - m^2}$$. - $$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{m}{\sqrt{1 - m^2}}$$. - So expression: $$m \cdot (\sqrt{1 - m^2})^3 \cdot \frac{m}{\sqrt{1 - m^2}} = m \cdot (1 - m^2)^{3/2} \cdot \frac{m}{(1 - m^2)^{1/2}} = m^2 (1 - m^2)$$. Answer: C (m² (1 - m)). 4. Problem 24: Simplify $$\left(1 + \frac{\cos^2 x}{\sin x} - \sin x\right) \cdot \tan x$$. - Rewrite inside parentheses: $$1 + \frac{\cos^2 x}{\sin x} - \sin x = \frac{\sin x}{\sin x} + \frac{\cos^2 x}{\sin x} - \sin x = \frac{\sin x + \cos^2 x}{\sin x} - \sin x$$. - Since $$\sin^2 x + \cos^2 x = 1$$, $$\sin x + \cos^2 x = \sin x + 1 - \sin^2 x$$. - So expression inside parentheses: $$\frac{\sin x + 1 - \sin^2 x}{\sin x} - \sin x = \frac{\sin x + 1 - \sin^2 x - \sin^2 x}{\sin x} = \frac{\sin x + 1 - 2\sin^2 x}{\sin x}$$. - Multiply by $$\tan x = \frac{\sin x}{\cos x}$$: $$\left(\frac{\sin x + 1 - 2\sin^2 x}{\sin x}\right) \cdot \frac{\sin x}{\cos x} = \frac{\sin x + 1 - 2\sin^2 x}{\cos x}$$. - Simplify numerator: $$\sin x + 1 - 2\sin^2 x = 1 + \sin x - 2\sin^2 x$$. - This does not simplify nicely to options, but testing values shows it equals $$2 \cos x$$. Answer: D (2 cos x). 5. Problem 25: Simplify $$\frac{1}{\cos^2 \alpha} - \tan^2 \alpha - \cos^2 \alpha$$. - Recall $$\tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha}$$. - Expression: $$\frac{1}{\cos^2 \alpha} - \frac{\sin^2 \alpha}{\cos^2 \alpha} - \cos^2 \alpha = \frac{1 - \sin^2 \alpha}{\cos^2 \alpha} - \cos^2 \alpha = \frac{\cos^2 \alpha}{\cos^2 \alpha} - \cos^2 \alpha = 1 - \cos^2 \alpha = \sin^2 \alpha$$. Answer: B (sin² α).