Given, `"sin" x = -(3)/(7)` and x belongs to the third quadrant, obtain the value of `"tan" x-sec x`
(ii) If `7 sin A = 24 cos A` and `0ltA lt(pi)/(2)` find`14 tan A - 25 cos A - 7 sec A`

Answer

To solve the given problems step by step, we will break down each part of the question.

### Part (i): Given \( \sin x = -\frac{3}{7} \) and \( x \) belongs to the third quadrant, find \( \tan x - \sec x \).

1. **Identify the values**:
- We know that \( \sin x = -\frac{3}{7} \). In the third quadrant, sine is negative.
- The hypotenuse (r) is 7, and the opposite side (perpendicular) is -3 (since we consider lengths, we take it as 3).

2. **Find the adjacent side**:
- Using the Pythagorean theorem:
\[ r^2 = \text{opposite}^2 + \text{adjacent}^2 \]
\[ 7^2 = 3^2 + \text{adjacent}^2 \]
\[ 49 = 9 + \text{adjacent}^2 \]
\[ \text{adjacent}^2 = 49 - 9 = 40 \]
\[ \text{adjacent} = \sqrt{40} = 2\sqrt{10} \]

3. **Calculate \( \tan x \)**:
- In the third quadrant, \( \tan x \) is positive:
\[ \tan x = \frac{\text{opposite}}{\text{adjacent}} = \frac{-3}{2\sqrt{10}} = -\frac{3}{2\sqrt{10}} \text{ (but we take the positive value)} = \frac{3}{2\sqrt{10}} \]

4. **Calculate \( \sec x \)**:
- \( \sec x = \frac{r}{\text{adjacent}} \):
\[ \sec x = \frac{7}{2\sqrt{10}} \text{ (but since we are in the third quadrant, it is negative)} = -\frac{7}{2\sqrt{10}} \]

5. **Find \( \tan x - \sec x \)**:
\[ \tan x - \sec x = \frac{3}{2\sqrt{10}} - \left(-\frac{7}{2\sqrt{10}}\right) = \frac{3}{2\sqrt{10}} + \frac{7}{2\sqrt{10}} = \frac{10}{2\sqrt{10}} = \frac{5}{\sqrt{10}} = \frac{5\sqrt{10}}{10} = \frac{\sqrt{10}}{2} \]

### Part (ii): Given \( 7 \sin A = 24 \cos A \) and \( 0 < A < \frac{\pi}{2} \), find \( 14 \tan A - 25 \cos A - 7 \sec A \).

1. **Express \( \tan A \)**:
\[ \tan A = \frac{\sin A}{\cos A} \Rightarrow \sin A = \frac{24}{7} \cos A \]

2. **Use Pythagorean identity**:
- From \( \sin^2 A + \cos^2 A = 1 \):
\[ \left(\frac{24}{7} \cos A\right)^2 + \cos^2 A = 1 \]
\[ \frac{576}{49} \cos^2 A + \cos^2 A = 1 \]
\[ \left(\frac{576}{49} + 1\right) \cos^2 A = 1 \]
\[ \left(\frac{576 + 49}{49}\right) \cos^2 A = 1 \]
\[ \frac{625}{49} \cos^2 A = 1 \Rightarrow \cos^2 A = \frac{49}{625} \Rightarrow \cos A = \frac{7}{25} \]

3. **Find \( \sin A \)**:
\[ \sin A = \frac{24}{7} \cos A = \frac{24}{7} \cdot \frac{7}{25} = \frac{24}{25} \]

4. **Calculate \( \tan A \)**:
\[ \tan A = \frac{\sin A}{\cos A} = \frac{\frac{24}{25}}{\frac{7}{25}} = \frac{24}{7} \]

5. **Calculate \( \sec A \)**:
\[ \sec A = \frac{1}{\cos A} = \frac{25}{7} \]

6. **Substitute values into the expression**:
\[ 14 \tan A - 25 \cos A - 7 \sec A = 14 \cdot \frac{24}{7} - 25 \cdot \frac{7}{25} - 7 \cdot \frac{25}{7} \]
\[ = 48 - 7 - 25 = 16 \]

### Final Answers:
- Part (i): \( \tan x - \sec x = \frac{\sqrt{10}}{2} \)
- Part (ii): \( 14 \tan A - 25 \cos A - 7 \sec A = 16 \)