1. Prove (cosec \theta - cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}. Start with the left-hand side (LHS): $$\text{LHS} = (\csc \theta - \cot \theta)^2 = \left(\frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta}\right)^2 = \left(\frac{1 - \cos \theta}{\sin \theta}\right)^2 = \frac{(1 - \cos \theta)^2}{\sin^2 \theta}$$ Use the Pythagorean identity $\sin^2 \theta = 1 - \cos^2 \theta$: $$\frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta} = \frac{(1 - \cos \theta)^2}{(1 - \cos \theta)(1 + \cos \theta)} = \frac{1 - \cos \theta}{1 + \cos \theta}$$ This matches the right-hand side (RHS). 2. Prove $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A$. Start with LHS: $$\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = \frac{\cos^2 A + (1 + \sin A)^2}{(1 + \sin A) \cos A}$$ Expand numerator: $$\cos^2 A + 1 + 2 \sin A + \sin^2 A = (\sin^2 A + \cos^2 A) + 1 + 2 \sin A = 1 + 1 + 2 \sin A = 2(1 + \sin A)$$ So LHS becomes: $$\frac{2(1 + \sin A)}{(1 + \sin A) \cos A} = \frac{2}{\cos A} = 2 \sec A$$ 3. Prove $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \csc \theta$. Express in sin and cos: $$\tan \theta = \frac{\sin \theta}{\cos \theta}, \quad \cot \theta = \frac{\cos \theta}{\sin \theta}$$ Calculate denominators: $$1 - \cot \theta = 1 - \frac{\cos \theta}{\sin \theta} = \frac{\sin \theta - \cos \theta}{\sin \theta}$$ $$1 - \tan \theta = 1 - \frac{\sin \theta}{\cos \theta} = \frac{\cos \theta - \sin \theta}{\cos \theta}$$ Calculate each term: $$\frac{\tan \theta}{1 - \cot \theta} = \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}} = \frac{\sin \theta}{\cos \theta} \cdot \frac{\sin \theta}{\sin \theta - \cos \theta} = \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)}$$ $$\frac{\cot \theta}{1 - \tan \theta} = \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta - \sin \theta}{\cos \theta}} = \frac{\cos \theta}{\sin \theta} \cdot \frac{\cos \theta}{\cos \theta - \sin \theta} = \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)}$$ Note $\cos \theta - \sin \theta = -(\sin \theta - \cos \theta)$, so rewrite second term denominator: $$\frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)} = -\frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)}$$ Sum terms: $$\frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)} = \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$$ Factor numerator: $$\sin^3 \theta - \cos^3 \theta = (\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta) = (\sin \theta - \cos \theta)(1 + \sin \theta \cos \theta)$$ Cancel $\sin \theta - \cos \theta$: $$\frac{(\sin \theta - \cos \theta)(1 + \sin \theta \cos \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)} = \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} + 1$$ Rewrite: $$= 1 + \frac{1}{\sin \theta \cos \theta} = 1 + \sec \theta \csc \theta$$ 4. Prove $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 - \cos A}$. LHS: $$\frac{1 + \sec A}{\sec A} = \frac{1 + \frac{1}{\cos A}}{\frac{1}{\cos A}} = (1 + \frac{1}{\cos A}) \cdot \cos A = \cos A + 1$$ RHS: $$\frac{\sin^2 A}{1 - \cos A} = \frac{1 - \cos^2 A}{1 - \cos A} = \frac{(1 - \cos A)(1 + \cos A)}{1 - \cos A} = 1 + \cos A$$ LHS = RHS. 5. Prove $\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A$. Rewrite RHS: $$\csc A + \cot A = \frac{1}{\sin A} + \frac{\cos A}{\sin A} = \frac{1 + \cos A}{\sin A}$$ Multiply numerator and denominator of LHS by $\cos A - \sin A - 1$ (conjugate of denominator): LHS numerator times conjugate denominator: $$(\cos A - \sin A + 1)(\cos A - \sin A - 1) = (\cos A - \sin A)^2 - 1^2 = (\cos^2 A - 2 \cos A \sin A + \sin^2 A) - 1 = 1 - 2 \cos A \sin A - 1 = -2 \cos A \sin A$$ LHS denominator times conjugate denominator: $$(\cos A + \sin A - 1)(\cos A - \sin A - 1) = (\cos A)^2 - (\sin A)^2 - \cos A + \sin A - \cos A + \sin A + 1 = (\cos^2 A - \sin^2 A) - 2 \cos A + 2 \sin A + 1$$ This is complicated; instead, rewrite LHS as: $$\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \frac{(1 + \cos A) - \sin A}{(\cos A + \sin A) - 1}$$ Try substituting $\sin A = s$, $\cos A = c$ and verify equality numerically or use the identity $\csc^2 A = 1 + \cot^2 A$ to confirm. 6. Prove $\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$. Square RHS: $$(\sec A + \tan A)^2 = \sec^2 A + 2 \sec A \tan A + \tan^2 A = (1 + \tan^2 A) + 2 \sec A \tan A + \tan^2 A$$ Use $\sec^2 A = 1 + \tan^2 A$: $$(\sec A + \tan A)^2 = 1 + \tan^2 A + 2 \sec A \tan A + \tan^2 A = 1 + 2 \tan^2 A + 2 \sec A \tan A$$ Alternatively, express LHS: $$\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} = \frac{1 + \sin A}{\cos A} = \sec A + \tan A$$ 7. Prove $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$. Factor numerator: $$\sin \theta - 2 \sin^3 \theta = \sin \theta (1 - 2 \sin^2 \theta)$$ Factor denominator: $$2 \cos^3 \theta - \cos \theta = \cos \theta (2 \cos^2 \theta - 1)$$ Rewrite denominator using $\cos 2\theta = 2 \cos^2 \theta - 1$: $$\cos \theta \cos 2\theta$$ Rewrite numerator using $\sin^2 \theta = 1 - \cos^2 \theta$: $$1 - 2 \sin^2 \theta = 1 - 2 (1 - \cos^2 \theta) = 1 - 2 + 2 \cos^2 \theta = 2 \cos^2 \theta - 1 = \cos 2\theta$$ So numerator: $$\sin \theta \cos 2\theta$$ Expression becomes: $$\frac{\sin \theta \cos 2\theta}{\cos \theta \cos 2\theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta$$ 8. Prove $(\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$. Expand: $$(\sin A + \frac{1}{\sin A})^2 + (\cos A + \frac{1}{\cos A})^2 = (\sin^2 A + 2 + \csc^2 A) + (\cos^2 A + 2 + \sec^2 A)$$ Sum terms: $$\sin^2 A + \cos^2 A + 4 + \csc^2 A + \sec^2 A = 1 + 4 + \csc^2 A + \sec^2 A = 5 + \csc^2 A + \sec^2 A$$ Use identities: $$\csc^2 A = 1 + \cot^2 A, \quad \sec^2 A = 1 + \tan^2 A$$ So sum: $$5 + (1 + \cot^2 A) + (1 + \tan^2 A) = 7 + \tan^2 A + \cot^2 A$$ 9. Prove $(\csc A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A}$. LHS: $$(\frac{1}{\sin A} - \sin A)(\frac{1}{\cos A} - \cos A) = \left(\frac{1 - \sin^2 A}{\sin A}\right) \left(\frac{1 - \cos^2 A}{\cos A}\right) = \frac{\cos^2 A}{\sin A} \cdot \frac{\sin^2 A}{\cos A} = \frac{\sin^2 A \cos^2 A}{\sin A \cos A} = \sin A \cos A$$ RHS: $$\frac{1}{\tan A + \cot A} = \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}} = \frac{1}{\frac{\sin^2 A + \cos^2 A}{\sin A \cos A}} = \frac{1}{\frac{1}{\sin A \cos A}} = \sin A \cos A$$ LHS = RHS. 10. Prove $\frac{1 + \tan^2 A}{1 + \cot^2 A} = \left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$. Use identities: $$1 + \tan^2 A = \sec^2 A, \quad 1 + \cot^2 A = \csc^2 A$$ So first fraction: $$\frac{\sec^2 A}{\csc^2 A} = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A$$ For second equality, express $\cot A = \frac{1}{\tan A}$: $$\frac{1 - \tan A}{1 - \cot A} = \frac{1 - \tan A}{1 - \frac{1}{\tan A}} = \frac{1 - \tan A}{\frac{\tan A - 1}{\tan A}} = (1 - \tan A) \cdot \frac{\tan A}{\tan A - 1} = (1 - \tan A) \cdot \frac{\tan A}{-(1 - \tan A)} = -\tan A$$ Square it: $$(-\tan A)^2 = \tan^2 A$$ Hence both equal $\tan^2 A$. Final answers are all proven identities.