Which of the following is (are) true?
I. `sin^(-1)1+sin^(-1)(-1)=0`
II. `cos^(-1)1+cos^(-1)(-1)=0`
III. `cos^(-1)x=cos^(-1)(-x)` for all x in the domain of `cos^(-1)`

Options

A. Only I
B. only II
C. only III
D. only I and II

Explanation

To determine which of the statements are true, we will analyze each one step by step.

### Step 1: Analyze Statement I
**Statement I:** \( \sin^{-1}(1) + \sin^{-1}(-1) = 0 \)

1. We know that \( \sin^{-1}(1) \) is the angle whose sine is 1. This angle is \( \frac{\pi}{2} \).
2. Similarly, \( \sin^{-1}(-1) \) is the angle whose sine is -1, which is \( -\frac{\pi}{2} \).
3. Now, we can add these two values:
\[ \sin^{-1}(1) + \sin^{-1}(-1) = \frac{\pi}{2} + \left(-\frac{\pi}{2}\right) = 0 \]
4. Therefore, Statement I is **true**.

### Step 2: Analyze Statement II
**Statement II:** \( \cos^{-1}(1) + \cos^{-1}(-1) = 0 \)

1. We know that \( \cos^{-1}(1) \) is the angle whose cosine is 1, which is \( 0 \).
2. \( \cos^{-1}(-1) \) is the angle whose cosine is -1, which is \( \pi \).
3. Now, we can add these two values:
\[ \cos^{-1}(1) + \cos^{-1}(-1) = 0 + \pi = \pi \]
4. Therefore, Statement II is **false**.

### Step 3: Analyze Statement III
**Statement III:** \( \cos^{-1}(x) = \cos^{-1}(-x) \) for all \( x \) in the domain of \( \cos^{-1} \)

1. The identity \( \cos^{-1}(-x) \) can be expressed as:
\[ \cos^{-1}(-x) = \pi - \cos^{-1}(x) \]
2. This shows that \( \cos^{-1}(x) \) and \( \cos^{-1}(-x) \) are not equal for all \( x \) in the domain of \( \cos^{-1} \).
3. For example, if \( x = 1 \):
\[ \cos^{-1}(1) = 0 \quad \text{and} \quad \cos^{-1}(-1) = \pi \]
4. Therefore, Statement III is **false**.

### Conclusion
- Statement I is true.
- Statement II is false.
- Statement III is false.

Thus, the only true statement is **I**.

---