Solve the trigonometry problems in the image.

Steps to Solve

Here are the solutions to problems 16, 18, 20, 24, 26 and 28:

1. Problem 16: Simplify the trigonometric expression Starting with the left-hand side (LHS): $$ \frac{1 + \tan^2{\theta}}{1 + \cot^2{\theta}} $$ Using the identities $1 + \tan^2{\theta} = \sec^2{\theta}$ and $1 + \cot^2{\theta} = \csc^2{\theta}$: $$ \frac{\sec^2{\theta}}{\csc^2{\theta}} $$ Since $\sec{\theta} = \frac{1}{\cos{\theta}}$ and $\csc{\theta} = \frac{1}{\sin{\theta}}$: $$ \frac{\frac{1}{\cos^2{\theta}}}{\frac{1}{\sin^2{\theta}}} = \frac{\sin^2{\theta}}{\cos^2{\theta}} = \tan^2{\theta} $$ Now, let's simplify the second part: $$ (\frac{1 - \tan{\theta}}{1 - \cot{\theta}})^2 = (\frac{1 - \tan{\theta}}{1 - \frac{1}{\tan{\theta}}})^2 = (\frac{1 - \tan{\theta}}{\frac{\tan{\theta} - 1}{\tan{\theta}}})^2 = (\frac{(1 - \tan{\theta})\tan{\theta}}{\tan{\theta} - 1})^2 = (-\tan{\theta})^2 = \tan^2{\theta} $$ Thus, $$ \frac{1 + \tan^2{\theta}}{1 + \cot^2{\theta}} = (\frac{1 - \tan{\theta}}{1 - \cot{\theta}})^2 = \tan^2{\theta} $$

2. Problem 18: Prove the trigonometric identity $$ \sec^6{\theta} = \tan^6{\theta} + 3\tan^2{\theta}\sec^2{\theta} + 1 $$ We know that $\sec^2{\theta} = 1 + \tan^2{\theta}$. Then $\sec^6{\theta} = (\sec^2{\theta})^3 = (1 + \tan^2{\theta})^3$. Using the binomial expansion $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$: $$ (1 + \tan^2{\theta})^3 = 1 + 3\tan^2{\theta} + 3\tan^4{\theta} + \tan^6{\theta} $$ We want to show this is equal to $\tan^6{\theta} + 3\tan^2{\theta}\sec^2{\theta} + 1$. Since $\sec^2{\theta} = 1 + \tan^2{\theta}$: $$ \tan^6{\theta} + 3\tan^2{\theta}(1 + \tan^2{\theta}) + 1 = \tan^6{\theta} + 3\tan^2{\theta} + 3\tan^4{\theta} + 1 $$ So we have: $$ \sec^6{\theta} = 1 + 3\tan^2{\theta} + 3\tan^4{\theta} + \tan^6{\theta} = \tan^6{\theta} + 3\tan^2{\theta}\sec^2{\theta} + 1 $$ Therefore, the identity is proven.

3. Problem 20: Simplify the trigonometric expression $$ (\sec{\theta} + \cos{\theta})(\sec{\theta} - \cos{\theta}) = \tan^2{\theta} + \sin^2{\theta} $$ Using the difference of squares formula $(a+b)(a-b) = a^2 - b^2$: $$ (\sec{\theta} + \cos{\theta})(\sec{\theta} - \cos{\theta}) = \sec^2{\theta} - \cos^2{\theta} $$ Since $\sec{\theta} = \frac{1}{\cos{\theta}}$: $$ \sec^2{\theta} - \cos^2{\theta} = \frac{1}{\cos^2{\theta}} - \cos^2{\theta} $$ We want to show this is equal to $\tan^2{\theta} + \sin^2{\theta}$. Using $\tan^2{\theta} = \frac{\sin^2{\theta}}{\cos^2{\theta}}$: $$ \tan^2{\theta} + \sin^2{\theta} = \frac{\sin^2{\theta}}{\cos^2{\theta}} + \sin^2{\theta} = \frac{\sin^2{\theta} + \sin^2{\theta}\cos^2{\theta}}{\cos^2{\theta}} = \frac{\sin^2{\theta}(1 + \cos^2{\theta})}{\cos^2{\theta}} $$ Using $\sin^2{\theta} = 1 - \cos^2{\theta}$: $$ \frac{1}{\cos^2{\theta}} - \cos^2{\theta} = \frac{1 - \cos^4{\theta}}{\cos^2{\theta}} = \frac{(1 - \cos^2{\theta})(1 + \cos^2{\theta})}{\cos^2{\theta}} = \frac{\sin^2{\theta}(1 + \cos^2{\theta})}{\cos^2{\theta}} = \frac{\sin^2{\theta}}{\cos^2{\theta}} + \sin^2{\theta} = \tan^2{\theta} + \sin^2{\theta} $$ Therefore, the identity is proven.

4. Problem 24: Simplify the trigonometric expression $$ \frac{1 + \cos{\theta} - \sin^2{\theta}}{\sin{\theta}(1 + \cos{\theta})} = \cot{\theta} $$ Since $\sin^2{\theta} = 1 - \cos^2{\theta}$: $$ \frac{1 + \cos{\theta} - (1 - \cos^2{\theta})}{\sin{\theta}(1 + \cos{\theta})} = \frac{\cos{\theta} + \cos^2{\theta}}{\sin{\theta}(1 + \cos{\theta})} = \frac{\cos{\theta}(1 + \cos{\theta})}{\sin{\theta}(1 + \cos{\theta})} = \frac{\cos{\theta}}{\sin{\theta}} = \cot{\theta} $$ Thus, the identity is proven.

5. Problem 26: Simplify the trigonometric expression $$ \frac{\tan A}{1 + \sec A} - \frac{\tan A}{1 - \sec A} = 2\csc A $$ $$ \frac{\tan A(1 - \sec A) - \tan A(1 + \sec A)}{(1 + \sec A)(1 - \sec A)} = \frac{\tan A - \tan A \sec A - \tan A - \tan A \sec A}{1 - \sec^2 A} = \frac{-2 \tan A \sec A}{1 - \sec^2 A} $$ Since $1 - \sec^2 A = -\tan^2 A$: $$ \frac{-2 \tan A \sec A}{-\tan^2 A} = \frac{2 \sec A}{\tan A} = \frac{2 \cdot \frac{1}{\cos A}}{\frac{\sin A}{\cos A}} = \frac{2}{\cos A} \cdot \frac{\cos A}{\sin A} = \frac{2}{\sin A} = 2\csc A $$ Thus, the identity is proven.

6. Problem 28: Simplify the trigonometric expression $$ \frac{\cos{\theta}}{\csc{\theta} + 1} + \frac{\cos{\theta}}{\csc{\theta} - 1} = 2 $$ $$ \frac{\cos{\theta}(\csc{\theta} - 1) + \cos{\theta}(\csc{\theta} + 1)}{(\csc{\theta} + 1)(\csc{\theta} - 1)} = \frac{\cos{\theta}\csc{\theta} - \cos{\theta} + \cos{\theta}\csc{\theta} + \cos{\theta}}{\csc^2{\theta} - 1} = \frac{2\cos{\theta}\csc{\theta}}{\csc^2{\theta} - 1} $$ Since $\csc{\theta} = \frac{1}{\sin{\theta}}$ and $\csc^2{\theta} - 1 = \cot^2{\theta}$: $$ \frac{2\cos{\theta}\cdot \frac{1}{\sin{\theta}}}{\cot^2{\theta}} = \frac{2\cot{\theta}}{\cot^2{\theta}} = \frac{2}{\cot{\theta}} = 2\tan{\theta} $$

It seems there's a mistake in the original problem. The correct identity should be: $$ \frac{\cos{\theta}}{\csc{\theta} + 1} + \frac{\cos{\theta}}{\csc{\theta} - 1} = 2\tan{\theta} $$ However, if we use the identity $\cot{\theta} = \frac{\cos{\theta}}{\sin{\theta}}$ then the last expression becomes:

$$ \frac{2\cos{\theta} \frac{1}{\sin{\theta}}}{\frac{\cos^2{\theta}}{\sin^2{\theta}}} = \frac{2 \frac{\cos{\theta}}{\sin{\theta}}}{\frac{\cos^2{\theta}}{\sin^2{\theta}}} = 2 \frac{\cos{\theta}}{\sin{\theta}} \frac{\sin^2{\theta}}{\cos^2{\theta}}= 2\frac{\sin{\theta}}{\cos{\theta}} = 2\tan{\theta} $$

There might be an additional error or typo somewhere because the expression does not equal 2.