1. Prove identity a) $\sec \theta - \cos \theta = \sin \theta \tan \theta$. Start with the left side (LHS): $$\sec \theta - \cos \theta = \frac{1}{\cos \theta} - \cos \theta$$ Find common denominator: $$= \frac{1 - \cos^2 \theta}{\cos \theta}$$ Use Pythagorean identity $1 - \cos^2 \theta = \sin^2 \theta$: $$= \frac{\sin^2 \theta}{\cos \theta}$$ Rewrite as: $$= \sin \theta \cdot \frac{\sin \theta}{\cos \theta} = \sin \theta \tan \theta$$ This equals the right side (RHS), so identity a) is proven. 2. Prove identity b) $\frac{\sin \theta \cos \theta}{1 + \cos \theta} = \frac{1 - \cos \theta}{\tan \theta}$. Start with RHS: $$\frac{1 - \cos \theta}{\tan \theta} = \frac{1 - \cos \theta}{\frac{\sin \theta}{\cos \theta}} = (1 - \cos \theta) \cdot \frac{\cos \theta}{\sin \theta}$$ Rewrite numerator and denominator: $$= \frac{(1 - \cos \theta) \cos \theta}{\sin \theta}$$ Multiply numerator and denominator of LHS by $1 - \cos \theta$: $$\frac{\sin \theta \cos \theta}{1 + \cos \theta} \cdot \frac{1 - \cos \theta}{1 - \cos \theta} = \frac{\sin \theta \cos \theta (1 - \cos \theta)}{1 - \cos^2 \theta}$$ Use $1 - \cos^2 \theta = \sin^2 \theta$: $$= \frac{\sin \theta \cos \theta (1 - \cos \theta)}{\sin^2 \theta} = \frac{\cos \theta (1 - \cos \theta)}{\sin \theta}$$ This matches RHS, so identity b) is proven. 3. Prove identity c) $\frac{2 \cos x + 2 \cos^2 x}{\sin 2x} = \frac{\sin x}{1 - \cos x}$. Rewrite numerator: $$2 \cos x + 2 \cos^2 x = 2 \cos x (1 + \cos x)$$ Rewrite denominator using double angle: $$\sin 2x = 2 \sin x \cos x$$ So LHS: $$\frac{2 \cos x (1 + \cos x)}{2 \sin x \cos x} = \frac{\cancel{2} \cancel{\cos x} (1 + \cos x)}{\cancel{2} \sin x \cancel{\cos x}} = \frac{1 + \cos x}{\sin x}$$ RHS: $$\frac{\sin x}{1 - \cos x}$$ Multiply numerator and denominator of RHS by $1 + \cos x$: $$\frac{\sin x (1 + \cos x)}{(1 - \cos x)(1 + \cos x)} = \frac{\sin x (1 + \cos x)}{1 - \cos^2 x}$$ Use $1 - \cos^2 x = \sin^2 x$: $$= \frac{\sin x (1 + \cos x)}{\sin^2 x} = \frac{1 + \cos x}{\sin x}$$ This equals LHS, so identity c) is proven. 4. Prove identity d) $\frac{\sec^2 \theta}{\sec^2 \theta - 1} = \csc^2 \theta$. Rewrite denominator using Pythagorean identity $\sec^2 \theta - 1 = \tan^2 \theta$: $$\frac{\sec^2 \theta}{\tan^2 \theta}$$ Rewrite in terms of sine and cosine: $$\sec^2 \theta = \frac{1}{\cos^2 \theta}, \quad \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}$$ So: $$\frac{\frac{1}{\cos^2 \theta}}{\frac{\sin^2 \theta}{\cos^2 \theta}} = \frac{1}{\cos^2 \theta} \cdot \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{1}{\sin^2 \theta} = \csc^2 \theta$$ Identity d) is proven. 5. Prove identity e) $\frac{\sin \varphi + \tan \varphi}{1 + \cos \varphi} = \frac{1}{\cot \varphi}$. Rewrite numerator: $$\sin \varphi + \tan \varphi = \sin \varphi + \frac{\sin \varphi}{\cos \varphi} = \sin \varphi \left(1 + \frac{1}{\cos \varphi}\right) = \sin \varphi \frac{1 + \cos \varphi}{\cos \varphi}$$ So LHS: $$\frac{\sin \varphi \frac{1 + \cos \varphi}{\cos \varphi}}{1 + \cos \varphi} = \frac{\sin \varphi}{\cos \varphi} = \tan \varphi$$ RHS: $$\frac{1}{\cot \varphi} = \tan \varphi$$ Identity e) is proven. 6. Prove identity f) $\frac{\sin 2x}{1 + \cos 2x} = \frac{\sec^2 x - 1}{\tan x}$. Rewrite numerator and denominator of LHS using double angle formulas: $$\sin 2x = 2 \sin x \cos x, \quad 1 + \cos 2x = 1 + (2 \cos^2 x - 1) = 2 \cos^2 x$$ So LHS: $$\frac{2 \sin x \cos x}{2 \cos^2 x} = \frac{\cancel{2} \sin x \cancel{\cos x}}{\cancel{2} \cos^2 x} = \frac{\sin x}{\cos x} = \tan x$$ Rewrite RHS: $$\sec^2 x - 1 = \tan^2 x$$ So RHS: $$\frac{\tan^2 x}{\tan x} = \tan x$$ Identity f) is proven. All identities a) to f) are proven successfully.