1. Problem: Find which expression is equivalent to $\sec^2 x$. Step 1: Recall the Pythagorean identity: $\sec^2 x = 1 + \tan^2 x$. Step 2: Therefore, $\sec^2 x$ is equivalent to $1 + \tan^2 x$, so among the options, $\tan^2 x$ is the closest related term. 2. Problem: Find the equivalent of $\frac{\sin x}{\csc x}$. Step 1: Recall that $\csc x = \frac{1}{\sin x}$. Step 2: Substitute: $\frac{\sin x}{\csc x} = \sin x \times \sin x = \sin^2 x$. 3. Problem: Identify which identity is not a Pythagorean identity. Step 1: Pythagorean identities are forms of $\sin^2 A + \cos^2 A = 1$, $1 + \cot^2 A = \csc^2 A$, and $\sec^2 A - \tan^2 A = 1$. Step 2: The expression $\tan A + \cot A$ is not a Pythagorean identity. 4. Problem: Find the correct trigonometric identity. Step 1: The correct identity is $\tan^2 A = \sec^2 A - 1$. 5. Problem: Find the reciprocal of sine. Step 1: The reciprocal of $\sin x$ is $\csc x$. 6. Problem: Express $\cot A$. Step 1: $\cot A = \frac{\cos A}{\sin A} = \frac{1}{\tan A}$. Step 2: So both b and c are correct. 7. Problem: Simplify $\frac{1 - \sin^2 x}{\cos^2 x}$. Step 1: Use $1 - \sin^2 x = \cos^2 x$. Step 2: Substitute: $\frac{\cos^2 x}{\cos^2 x} = 1$. 8. Problem: Given $\tan x = \frac{3}{4}$, find $\sin x$. Step 1: Use right triangle ratios: opposite = 3, adjacent = 4. Step 2: Hypotenuse $= \sqrt{3^2 + 4^2} = 5$. Step 3: $\sin x = \frac{3}{5}$. 9. Problem: Simplify $\sec x$. Step 1: $\sec x = \frac{1}{\cos x}$, so it is not equal to 1, 0, $\sin x$, or $\tan x$. Step 2: None of the options except 1 is correct if $\cos x = 1$; otherwise, $\sec x$ is $\frac{1}{\cos x}$. 10. Problem: Simplify $\sec A \cot A \sin A$. Step 1: Write in terms of sine and cosine: $\sec A = \frac{1}{\cos A}$, $\cot A = \frac{\cos A}{\sin A}$. Step 2: Multiply: $\frac{1}{\cos A} \times \frac{\cos A}{\sin A} \times \sin A = 1$. 11. Problem: Simplify $\cot^2 A (1 + \tan^2 A)$. Step 1: Use identity $1 + \tan^2 A = \sec^2 A$. Step 2: Expression becomes $\cot^2 A \sec^2 A$. Step 3: $\cot A = \frac{\cos A}{\sin A}$ and $\sec A = \frac{1}{\cos A}$. Step 4: So $\cot^2 A \sec^2 A = \frac{\cos^2 A}{\sin^2 A} \times \frac{1}{\cos^2 A} = \frac{1}{\sin^2 A} = \csc^2 A$. 12. Problem: Identify an equation true for all valid variable replacements. Step 1: Such equations are called trigonometric identities. 13. Problem: Write $\cos^2 x \tan^2 x$ as a single term. Step 1: Recall $\tan x = \frac{\sin x}{\cos x}$. Step 2: Substitute: $\cos^2 x \times \frac{\sin^2 x}{\cos^2 x} = \sin^2 x$. 14. Problem: Simplify $\frac{\csc^2 x}{\sec^2 x}$. Step 1: $\csc^2 x = \frac{1}{\sin^2 x}$, $\sec^2 x = \frac{1}{\cos^2 x}$. Step 2: So $\frac{\csc^2 x}{\sec^2 x} = \frac{1/\sin^2 x}{1/\cos^2 x} = \frac{\cos^2 x}{\sin^2 x} = \cot^2 x$. 15. Problem: Simplify $\cot x \cos x \tan x \csc x$. Step 1: Express all in sine and cosine: $\cot x = \frac{\cos x}{\sin x}$, $\tan x = \frac{\sin x}{\cos x}$, $\csc x = \frac{1}{\sin x}$. Step 2: Multiply: $\frac{\cos x}{\sin x} \times \cos x \times \frac{\sin x}{\cos x} \times \frac{1}{\sin x} = \frac{\cos x}{\sin x} \times \cos x \times \frac{\sin x}{\cos x} \times \frac{1}{\sin x}$. Step 3: Simplify stepwise: $= \frac{\cos x}{\sin x} \times \cos x \times \frac{\sin x}{\cos x} \times \frac{1}{\sin x} = \frac{\cos x}{\sin x} \times \cos x \times \frac{1}{\cos x} \times \frac{1}{\sin x} = \frac{\cos x}{\sin x} \times 1 \times \frac{1}{\sin x} = \frac{\cos x}{\sin^2 x}$. Step 4: This does not simplify to any of the options directly; re-check. Step 5: Alternatively, cancel terms: $\cot x \times \tan x = 1$, so expression reduces to $1 \times \cos x \times \csc x = \cos x \times \frac{1}{\sin x} = \cot x$. 16. Problem: Simplify $\cos q (\csc q - \sec q) + 1$. Step 1: Write $\csc q = \frac{1}{\sin q}$ and $\sec q = \frac{1}{\cos q}$. Step 2: Expression becomes $\cos q \left( \frac{1}{\sin q} - \frac{1}{\cos q} \right) + 1 = \cos q \left( \frac{\cos q - \sin q}{\sin q \cos q} \right) + 1$. Step 3: Simplify numerator and denominator: $= \frac{\cos q (\cos q - \sin q)}{\sin q \cos q} + 1 = \frac{\cos q - \sin q}{\sin q} + 1$. Step 4: Combine terms: $= \frac{\cos q - \sin q + \sin q}{\sin q} = \frac{\cos q}{\sin q} = \cot q$. 17. Problem: Find an equivalent expression for $\csc A$. Step 1: $\csc A = \frac{1}{\sin A}$. 18. Problem: Simplify $\frac{\cos x}{1 + \sin x}$. Step 1: Multiply numerator and denominator by $1 - \sin x$ to rationalize denominator: $\frac{\cos x}{1 + \sin x} \times \frac{1 - \sin x}{1 - \sin x} = \frac{\cos x (1 - \sin x)}{1 - \sin^2 x}$. Step 2: Use $1 - \sin^2 x = \cos^2 x$: $= \frac{\cos x (1 - \sin x)}{\cos^2 x} = \frac{1 - \sin x}{\cos x}$. 19. Problem: Simplify $\tan q + \cot q$. Step 1: Write in sine and cosine: $\tan q = \frac{\sin q}{\cos q}$, $\cot q = \frac{\cos q}{\sin q}$. Step 2: Sum: $\frac{\sin q}{\cos q} + \frac{\cos q}{\sin q} = \frac{\sin^2 q + \cos^2 q}{\sin q \cos q} = \frac{1}{\sin q \cos q}$. 20. Problem: Simplify $\tan^2 q - \sin^2 q$. Step 1: Write $\tan^2 q = \frac{\sin^2 q}{\cos^2 q}$. Step 2: Expression: $\frac{\sin^2 q}{\cos^2 q} - \sin^2 q = \sin^2 q \left( \frac{1}{\cos^2 q} - 1 \right) = \sin^2 q \frac{1 - \cos^2 q}{\cos^2 q} = \sin^2 q \frac{\sin^2 q}{\cos^2 q} = \frac{\sin^4 q}{\cos^2 q}$. 21. Problem: Simplify $\csc^2 q \cos^2 q$. Step 1: $\csc^2 q = \frac{1}{\sin^2 q}$. Step 2: Expression: $\frac{1}{\sin^2 q} \times \cos^2 q = \frac{\cos^2 q}{\sin^2 q} = \cot^2 q$. 22. Problem: Simplify $(\sin q + \cos q)^2$. Step 1: Expand: $\sin^2 q + 2 \sin q \cos q + \cos^2 q$. Step 2: Use $\sin^2 q + \cos^2 q = 1$: $1 + 2 \sin q \cos q$. 23. Problem: Simplify $1 - 2 \sin^2 q$. Step 1: Use identity $\cos 2q = 1 - 2 \sin^2 q$. Step 2: So $1 - 2 \sin^2 q = \cos 2q$. Step 3: Among options, $2 \cos^2 q - 1$ is equivalent to $\cos 2q$. 24. Problem: Simplify $\csc x - \frac{1}{\csc x}$. Step 1: Write $\csc x = \frac{1}{\sin x}$. Step 2: Expression: $\frac{1}{\sin x} - \sin x = \frac{1 - \sin^2 x}{\sin x} = \frac{\cos^2 x}{\sin x}$. Step 3: This is not among options; re-check. Step 4: Alternatively, $\csc x - \frac{1}{\csc x} = \csc x - \sin x$. Step 5: Substitute $\csc x = \frac{1}{\sin x}$: $\frac{1}{\sin x} - \sin x = \frac{1 - \sin^2 x}{\sin x} = \frac{\cos^2 x}{\sin x}$. Step 6: None of the options match exactly; closest is $\sin x$ or $-\sin x$. 25. Problem: Simplify $(1 + \sin x)(1 - \sin x)$. Step 1: Use difference of squares: $1 - \sin^2 x = \cos^2 x$. Final answers: 1. $\tan^2 x$ 2. $\sin^2 x$ 3. $\tan A + \cot A$ 4. $\tan^2 A = \sec^2 A - 1$ 5. $\csc x$ 6. both b and c 7. $1$ 8. $\frac{3}{5}$ 9. $\frac{1}{\cos x}$ (none of the options exactly) 10. $1$ 11. $\csc^2 A$ 12. Trigonometric Identities 13. $\sin^2 x$ 14. $\cot^2 x$ 15. $\cot x$ 16. $\cot q$ 17. $\frac{1}{\sin A}$ 18. $\frac{1 - \sin x}{\cos x}$ 19. $\frac{1}{\sin q \cos q}$ 20. $\frac{\sin^4 q}{\cos^2 q}$ 21. $\cot^2 q$ 22. $1 + 2 \sin q \cos q$ 23. $2 \cos^2 q - 1$ 24. $\frac{\cos^2 x}{\sin x}$ (none of the options exactly) 25. $\cos^2 x$