(i) Evaluate : `sec (cos^(-1).(1)/(2))`
(ii) slove the equations ` sin^(-1) x + sin^(-1) y = (2pi)/(3)` and ` cos^(-1) x - cos ^(-1) y = (pi)/(3)`

Answer

### Step-by-Step Solution

#### Part (i): Evaluate \( \sec(\cos^{-1}(1/2)) \)

1. **Define the angle**: Let \( \theta = \cos^{-1}(1/2) \). This means that \( \cos(\theta) = 1/2 \).

**Hint**: Remember that \( \cos^{-1}(x) \) gives you the angle whose cosine is \( x \).

2. **Identify the triangle**: In a right triangle, if \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \), we can set the adjacent side to 1 and the hypotenuse to 2.

3. **Calculate the opposite side**: Using the Pythagorean theorem:
\[ \text{opposite} = \sqrt{\text{hypotenuse}^2 - \text{adjacent}^2} = \sqrt{2^2 - 1^2} = \sqrt{4 - 1} = \sqrt{3} \]

4. **Find \( \sec(\theta) \)**: Recall that \( \sec(\theta) = \frac{1}{\cos(\theta)} \). Since \( \cos(\theta) = 1/2 \):
\[ \sec(\theta) = \frac{1}{1/2} = 2 \]

5. **Final result**: Therefore, \( \sec(\cos^{-1}(1/2)) = 2 \).

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#### Part (ii): Solve the equations \( \sin^{-1}(x) + \sin^{-1}(y) = \frac{2\pi}{3} \) and \( \cos^{-1}(x) - \cos^{-1}(y) = \frac{\pi}{3} \)

1. **Label the equations**: Let:
- Equation (1): \( \sin^{-1}(x) + \sin^{-1}(y) = \frac{2\pi}{3} \)
- Equation (2): \( \cos^{-1}(x) - \cos^{-1}(y) = \frac{\pi}{3} \)

2. **Add the equations**: Adding Equation (1) and Equation (2):
\[ \sin^{-1}(x) + \cos^{-1}(x) + \sin^{-1}(y) - \cos^{-1}(y) = \frac{2\pi}{3} + \frac{\pi}{3} \]
Using the identity \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \):
\[ \frac{\pi}{2} + \sin^{-1}(y) - \cos^{-1}(y) = \pi \]

3. **Simplify**: Rearranging gives:
\[ \sin^{-1}(y) - \cos^{-1}(y) = \pi - \frac{\pi}{2} = \frac{\pi}{2} \]

4. **Use the identity**: From the identity \( \sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2} \), we can express:
\[ \sin^{-1}(y) = \frac{\pi}{2} + \cos^{-1}(y) \]
Substituting into the previous equation:
\[ \frac{\pi}{2} + \cos^{-1}(y) - \cos^{-1}(y) = \frac{\pi}{2} \]

5. **Solve for \( y \)**: This implies that \( y = 1 \) (since \( \sin^{-1}(1) = \frac{\pi}{2} \)).

6. **Substitute \( y \) back**: Now substitute \( y = 1 \) into Equation (1):
\[ \sin^{-1}(x) + \sin^{-1}(1) = \frac{2\pi}{3} \]
Therefore:
\[ \sin^{-1}(x) + \frac{\pi}{2} = \frac{2\pi}{3} \]

7. **Solve for \( x \)**: Rearranging gives:
\[ \sin^{-1}(x) = \frac{2\pi}{3} - \frac{\pi}{2} = \frac{4\pi - 3\pi}{6} = \frac{\pi}{6} \]
Thus:
\[ x = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]

8. **Final results**: The solutions are \( x = \frac{1}{2} \) and \( y = 1 \).

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