1. Problem 34: If $\log_2 x = \log_2 25$, find $x$. Since logarithm base 2 of $x$ equals logarithm base 2 of 25, by the property of logarithms, the arguments must be equal: $$x = 25$$ Answer: $x = 25$ (not among options a) 1, b) 2, c) 5, d) 3). Possibly a typo or missing option. 2. Problem 25: Simplify $\frac{\tan \theta \cot \theta}{\sec \theta}$. Recall: - $\tan \theta = \frac{\sin \theta}{\cos \theta}$ - $\cot \theta = \frac{\cos \theta}{\sin \theta}$ - $\sec \theta = \frac{1}{\cos \theta}$ Calculate numerator: $$\tan \theta \cot \theta = \frac{\sin \theta}{\cos \theta} \times \frac{\cos \theta}{\sin \theta} = 1$$ Divide by denominator: $$\frac{1}{\sec \theta} = \cos \theta$$ Answer: $\cos \theta$ (option c). 3. Problem 26: Simplify $\sin^2 \theta + \cos^2 \theta + \tan^2 \theta$. Recall: - $\sin^2 \theta + \cos^2 \theta = 1$ - $\tan^2 \theta = \sec^2 \theta - 1$ Substitute: $$1 + (\sec^2 \theta - 1) = \sec^2 \theta$$ Answer: $\sec^2 \theta$ (option c). 4. Problem 27: Simplify $2 \sin^2 \theta + \cos^2 \theta + \frac{1}{\sec^2 \theta}$. Recall: - $\frac{1}{\sec^2 \theta} = \cos^2 \theta$ - $\sin^2 \theta + \cos^2 \theta = 1$ Rewrite: $$2 \sin^2 \theta + \cos^2 \theta + \cos^2 \theta = 2 \sin^2 \theta + 2 \cos^2 \theta = 2(\sin^2 \theta + \cos^2 \theta) = 2 \times 1 = 2$$ Answer: 2 (option a). 5. Problem 28: Simplify $5 \sin \theta \csc \theta + 2 \cos \theta \sec \theta + 3 \tan \theta \cot \theta$. Recall: - $\csc \theta = \frac{1}{\sin \theta}$ - $\sec \theta = \frac{1}{\cos \theta}$ - $\tan \theta \cot \theta = 1$ Calculate each term: $$5 \sin \theta \times \frac{1}{\sin \theta} = 5$$ $$2 \cos \theta \times \frac{1}{\cos \theta} = 2$$ $$3 \times 1 = 3$$ Sum: $$5 + 2 + 3 = 10$$ Answer: 10 (option a). 6. Problem 29: Simplify $\sin^2 x (1 + \cot^2 x)$. Recall: - $1 + \cot^2 x = \csc^2 x$ - $\csc x = \frac{1}{\sin x}$ Rewrite: $$\sin^2 x \times \csc^2 x = \sin^2 x \times \frac{1}{\sin^2 x} = 1$$ Answer: 1 (option a). 7. Problem 30: Simplify $\sec \theta - \sin \theta \tan \theta$. Recall: - $\sec \theta = \frac{1}{\cos \theta}$ - $\tan \theta = \frac{\sin \theta}{\cos \theta}$ Calculate second term: $$\sin \theta \times \frac{\sin \theta}{\cos \theta} = \frac{\sin^2 \theta}{\cos \theta}$$ Rewrite expression: $$\frac{1}{\cos \theta} - \frac{\sin^2 \theta}{\cos \theta} = \frac{1 - \sin^2 \theta}{\cos \theta}$$ Recall $1 - \sin^2 \theta = \cos^2 \theta$: $$\frac{\cos^2 \theta}{\cos \theta} = \cos \theta$$ Answer: $\cos \theta$ (option b). 8. Problem 31: Simplify $\frac{1}{1 - \sin^2 x} - \frac{1}{1 - \cos^2 x}$. Recall: - $1 - \sin^2 x = \cos^2 x$ - $1 - \cos^2 x = \sin^2 x$ Rewrite: $$\frac{1}{\cos^2 x} - \frac{1}{\sin^2 x} = \sec^2 x - \csc^2 x$$ Recall identity: $$\csc^2 x - \cot^2 x = 1$$ But here, difference is $\sec^2 x - \csc^2 x$, no direct simplification to options. Check options: a) $\cot^2 \theta$ b) $\csc^2 \theta$ c) $\sec^2 \theta$ d) 1 No direct match; expression equals $\sec^2 x - \csc^2 x$. 9. Problem 32: Simplify $\frac{\sec^2 x - \tan^2 x}{1 + \tan^2 x + \cos^2 \theta}$. Recall: - $\sec^2 x - \tan^2 x = 1$ - $1 + \tan^2 x = \sec^2 x$ Rewrite denominator: $$\sec^2 x + \cos^2 \theta$$ No further simplification without values. Answer is $\frac{1}{\sec^2 x + \cos^2 \theta}$, no option matches. 10. Problem 33: Simplify $\frac{1 + \tan^2 \theta}{1 + \cot^2 \theta}$. Recall: - $1 + \tan^2 \theta = \sec^2 \theta$ - $1 + \cot^2 \theta = \csc^2 \theta$ Rewrite: $$\frac{\sec^2 \theta}{\csc^2 \theta} = \sec^2 \theta \times \sin^2 \theta = \frac{1}{\cos^2 \theta} \times \sin^2 \theta = \tan^2 \theta$$ Answer: $\tan^2 \theta$ (option c). 11. Problem 34 (second): Simplify $\frac{\sin \theta \cos \theta \tan \theta + \sin \theta \cos \theta \cot \theta}{\sin \theta \sec \theta}$. Recall: - $\tan \theta = \frac{\sin \theta}{\cos \theta}$ - $\cot \theta = \frac{\cos \theta}{\sin \theta}$ - $\sec \theta = \frac{1}{\cos \theta}$ Calculate numerator: $$\sin \theta \cos \theta \times \frac{\sin \theta}{\cos \theta} + \sin \theta \cos \theta \times \frac{\cos \theta}{\sin \theta} = \sin^2 \theta + \cos^2 \theta = 1$$ Calculate denominator: $$\sin \theta \times \frac{1}{\cos \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta$$ Expression: $$\frac{1}{\tan \theta} = \cot \theta$$ Answer: $\cot \theta$ (not among options). Possibly missing option. Summary of answers: 34) $x=25$ (not in options) 25) $\cos \theta$ (c) 26) $\sec^2 \theta$ (c) 27) 2 (a) 28) 10 (a) 29) 1 (a) 30) $\cos \theta$ (b) 31) Expression equals $\sec^2 x - \csc^2 x$ (no direct option) 32) $\frac{1}{\sec^2 x + \cos^2 \theta}$ (no direct option) 33) $\tan^2 \theta$ (c) 34 second) $\cot \theta$ (no option)