Find the value of the remainig(all) function in each of the following :
(i) `"sin"theta theta =(3)/(5), theta` in quadrient II
(ii) `costheta=-(3)/(5),theta` in quadrient III
(iii) `"tan" theta =(4)/(3), theta` in quadrant III
(iv) `cot theta=-(5)/(12),theta` in quadrant IV
(vi) `"sin" theta sectheta =- 1, theta` in quadant II.

Answer

To solve the given trigonometric problems step by step, we will use the fundamental relationships between the trigonometric functions and the properties of triangles.

### (i) Given: \( \sin \theta = \frac{3}{5} \), \( \theta \) in Quadrant II

1. **Find \( \cos \theta \)**:
- Use the Pythagorean identity: \( \sin^2 \theta + \cos^2 \theta = 1 \)
- Substitute \( \sin \theta \):
\[ \left(\frac{3}{5}\right)^2 + \cos^2 \theta = 1 \]
\[ \frac{9}{25} + \cos^2 \theta = 1 \]
\[ \cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25} \]
\[ \cos \theta = -\frac{4}{5} \quad (\text{negative in Quadrant II}) \]

2. **Find \( \tan \theta \)**:
- \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)
\[ \tan \theta = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4} \]

3. **Find \( \cot \theta \)**:
- \( \cot \theta = \frac{1}{\tan \theta} = -\frac{4}{3} \)

4. **Find \( \sec \theta \)**:
- \( \sec \theta = \frac{1}{\cos \theta} = -\frac{5}{4} \)

5. **Find \( \csc \theta \)**:
- \( \csc \theta = \frac{1}{\sin \theta} = \frac{5}{3} \)

### Summary for (i):
- \( \cos \theta = -\frac{4}{5} \)
- \( \tan \theta = -\frac{3}{4} \)
- \( \cot \theta = -\frac{4}{3} \)
- \( \sec \theta = -\frac{5}{4} \)
- \( \csc \theta = \frac{5}{3} \)

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### (ii) Given: \( \cos \theta = -\frac{3}{5} \), \( \theta \) in Quadrant III

1. **Find \( \sin \theta \)**:
- Use the Pythagorean identity:
\[ \sin^2 \theta + \left(-\frac{3}{5}\right)^2 = 1 \]
\[ \sin^2 \theta + \frac{9}{25} = 1 \]
\[ \sin^2 \theta = 1 - \frac{9}{25} = \frac{16}{25} \]
\[ \sin \theta = -\frac{4}{5} \quad (\text{negative in Quadrant III}) \]

2. **Find \( \tan \theta \)**:
- \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)
\[ \tan \theta = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{4}{3} \]

3. **Find \( \cot \theta \)**:
- \( \cot \theta = \frac{1}{\tan \theta} = \frac{3}{4} \)

4. **Find \( \sec \theta \)**:
- \( \sec \theta = \frac{1}{\cos \theta} = -\frac{5}{3} \)

5. **Find \( \csc \theta \)**:
- \( \csc \theta = \frac{1}{\sin \theta} = -\frac{5}{4} \)

### Summary for (ii):
- \( \sin \theta = -\frac{4}{5} \)
- \( \tan \theta = \frac{4}{3} \)
- \( \cot \theta = \frac{3}{4} \)
- \( \sec \theta = -\frac{5}{3} \)
- \( \csc \theta = -\frac{5}{4} \)

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### (iii) Given: \( \tan \theta = \frac{4}{3} \), \( \theta \) in Quadrant III

1. **Find \( \sin \theta \)**:
- Use \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and the identity \( \sin^2 \theta + \cos^2 \theta = 1 \).
- Let \( \sin \theta = 4k \) and \( \cos \theta = -3k \) (since \( \tan \theta = \frac{4}{3} \)).
\[ (4k)^2 + (-3k)^2 = 1 \]
\[ 16k^2 + 9k^2 = 1 \]
\[ 25k^2 = 1 \quad \Rightarrow \quad k^2 = \frac{1}{25} \quad \Rightarrow \quad k = \frac{1}{5} \]
\[ \sin \theta = 4k = \frac{4}{5}, \quad \cos \theta = -3k = -\frac{3}{5} \]

2. **Find \( \cot \theta \)**:
- \( \cot \theta = \frac{1}{\tan \theta} = \frac{3}{4} \)

3. **Find \( \sec \theta \)**:
- \( \sec \theta = \frac{1}{\cos \theta} = -\frac{5}{3} \)

4. **Find \( \csc \theta \)**:
- \( \csc \theta = \frac{1}{\sin \theta} = \frac{5}{4} \)

### Summary for (iii):
- \( \sin \theta = -\frac{4}{5} \)
- \( \tan \theta = \frac{4}{3} \)
- \( \cot \theta = \frac{3}{4} \)
- \( \sec \theta = -\frac{5}{3} \)
- \( \csc \theta = -\frac{5}{4} \)

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### (iv) Given: \( \cot \theta = -\frac{5}{12} \), \( \theta \) in Quadrant IV

1. **Find \( \tan \theta \)**:
- \( \tan \theta = -\frac{12}{5} \)

2. **Find \( \sin \theta \)**:
- Let \( \sin \theta = -12k \) and \( \cos \theta = 5k \) (since \( \cot \theta = \frac{1}{\tan \theta} \)).
\[ (-12k)^2 + (5k)^2 = 1 \]
\[ 144k^2 + 25k^2 = 1 \]
\[ 169k^2 = 1 \quad \Rightarrow \quad k^2 = \frac{1}{169} \quad \Rightarrow \quad k = \frac{1}{13} \]
\[ \sin \theta = -\frac{12}{13}, \quad \cos \theta = \frac{5}{13} \]

3. **Find \( \sec \theta \)**:
- \( \sec \theta = \frac{1}{\cos \theta} = \frac{13}{5} \)

4. **Find \( \csc \theta \)**:
- \( \csc \theta = \frac{1}{\sin \theta} = -\frac{13}{12} \)

### Summary for (iv):
- \( \sin \theta = -\frac{12}{13} \)
- \( \tan \theta = -\frac{12}{5} \)
- \( \cot \theta = -\frac{5}{12} \)
- \( \sec \theta = \frac{13}{5} \)
- \( \csc \theta = -\frac{13}{12} \)

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### (v) Given: \( \sin \theta \sec \theta = -1 \), \( \theta \) in Quadrant II

1. **Use the identity**:
- \( \sec \theta = \frac{1}{\cos \theta} \)
- Thus, \( \sin \theta \cdot \frac{1}{\cos \theta} = -1 \)
- Rearranging gives \( \sin \theta = -\cos \theta \)

2. **Find \( \sin \theta \)**:
- Since \( \theta \) is in Quadrant II, \( \sin \theta \) is positive and \( \cos \theta \) is negative.
- Let \( \sin \theta = k \) and \( \cos \theta = -k \).
- Then \( k = -(-k) = k \) leads to \( k = \frac{1}{\sqrt{2}} \).

3. **Find \( \cos \theta \)**:
- \( \cos \theta = -\frac{1}{\sqrt{2}} \)

4. **Find \( \tan \theta \)**:
- \( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}} = -1 \)

5. **Find \( \cot \theta \)**:
- \( \cot \theta = -1 \)

6. **Find \( \sec \theta \)**:
- \( \sec \theta = -\sqrt{2} \)

7. **Find \( \csc \theta \)**:
- \( \csc \theta = \sqrt{2} \)

### Summary for (v):
- \( \sin \theta = \frac{1}{\sqrt{2}} \)
- \( \tan \theta = -1 \)
- \( \cot \theta = -1 \)
- \( \sec \theta = -\sqrt{2} \)
- \( \csc \theta = \sqrt{2} \)

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