In the following figure :

`AD bot BC, AC = 26, CD = 10, BC = 42, angle DAC = x and angle B = y`.
Find the value of :
(i) cot x
(ii) `(1)/(sin^(2)y) - (1)/(tan^(2)y)`
(iii) `(6)/(cos x) - (5)/(cos y) + 8 tan y`

Answer

To solve the given problem step by step, we will break it down into three parts as specified in the question.

### Given:
- AC = 26
- CD = 10
- BC = 42
- Angle DAC = x
- Angle B = y

### Step 1: Finding cot x

1. **Identify the triangle ADC**:
- In triangle ADC, AD is perpendicular to BC.
- We know AC (hypotenuse) = 26 and CD (perpendicular) = 10.

2. **Calculate the length of AD using Pythagoras theorem**:
- According to the Pythagorean theorem:
\[ AC^2 = AD^2 + CD^2 \]
- Plugging in the values:
\[ 26^2 = AD^2 + 10^2 \]
\[ 676 = AD^2 + 100 \]
\[ AD^2 = 676 - 100 = 576 \]
\[ AD = \sqrt{576} = 24 \]

3. **Calculate cot x**:
- Cotangent is defined as the adjacent side over the opposite side.
- In triangle ADC, cot x = AD/CD.
- Thus,
\[ \cot x = \frac{AD}{CD} = \frac{24}{10} = 2.4 \]

### Step 2: Finding \( \frac{1}{\sin^2 y} - \frac{1}{\tan^2 y} \)

1. **Identify triangle ADB**:
- In triangle ADB, we have:
- AD = 24 (perpendicular)
- BD = BC - CD = 42 - 10 = 32 (base)

2. **Calculate AB using Pythagorean theorem**:
- Using the Pythagorean theorem:
\[ AB^2 = AD^2 + BD^2 \]
- Plugging in the values:
\[ AB^2 = 24^2 + 32^2 = 576 + 1024 = 1600 \]
\[ AB = \sqrt{1600} = 40 \]

3. **Calculate \( \sin y \) and \( \tan y \)**:
- \( \sin y = \frac{AD}{AB} = \frac{24}{40} = \frac{3}{5} \)
- \( \tan y = \frac{AD}{BD} = \frac{24}{32} = \frac{3}{4} \)

4. **Calculate \( \frac{1}{\sin^2 y} - \frac{1}{\tan^2 y} \)**:
- \( \sin^2 y = \left(\frac{3}{5}\right)^2 = \frac{9}{25} \)
- \( \tan^2 y = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \)
- Thus,
\[ \frac{1}{\sin^2 y} = \frac{25}{9}, \quad \frac{1}{\tan^2 y} = \frac{16}{9} \]
- Therefore,
\[ \frac{1}{\sin^2 y} - \frac{1}{\tan^2 y} = \frac{25}{9} - \frac{16}{9} = \frac{9}{9} = 1 \]

### Step 3: Finding \( \frac{6}{\cos x} - \frac{5}{\cos y} + 8 \tan y \)

1. **Calculate \( \cos x \) and \( \cos y \)**:
- For \( \cos x \):
\[ \cos x = \frac{AD}{AC} = \frac{24}{26} = \frac{12}{13} \]
- For \( \cos y \):
\[ \cos y = \frac{BD}{AB} = \frac{32}{40} = \frac{4}{5} \]

2. **Substituting values into the expression**:
- Calculate \( \frac{6}{\cos x} \):
\[ \frac{6}{\cos x} = \frac{6}{\frac{12}{13}} = 6 \cdot \frac{13}{12} = \frac{78}{12} = 6.5 \]
- Calculate \( \frac{5}{\cos y} \):
\[ \frac{5}{\cos y} = \frac{5}{\frac{4}{5}} = 5 \cdot \frac{5}{4} = \frac{25}{4} = 6.25 \]
- Calculate \( 8 \tan y \):
\[ 8 \tan y = 8 \cdot \frac{3}{4} = 6 \]

3. **Final calculation**:
- Combine all parts:
\[ \frac{6}{\cos x} - \frac{5}{\cos y} + 8 \tan y = 6.5 - 6.25 + 6 = 6.25 + 6 = 12.25 \]

### Final Answers:
(i) \( \cot x = 2.4 \)
(ii) \( \frac{1}{\sin^2 y} - \frac{1}{\tan^2 y} = 1 \)
(iii) \( \frac{6}{\cos x} - \frac{5}{\cos y} + 8 \tan y = 12.25 \)