If x takes negative permissible value then `sin^(-1)x=`

Options

A. `cos^(-1)sqrt(1-x^(2))`
B. `-cos^(-1)sqrt(1-x^(2))`
C. `cos^(-1)sqrt(x^(2)-1)`
D. `pi-cos^(-1)sqrt(1-x^(2))`

Answer

To solve the problem, we need to find the expression for \(\sin^{-1}(x)\) when \(x\) takes negative permissible values.

### Step-by-Step Solution:

1. **Understanding the Range of \(\sin^{-1}(x)\)**:
The function \(\sin^{-1}(x)\) is defined for \(x\) in the range \([-1, 1]\). When \(x\) takes negative values, we are looking at the interval \([-1, 0)\).

2. **Using the Relationship with \(\cos^{-1}\)**:
We know that:
\[ \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \]
This relationship can be rearranged to express \(\sin^{-1}(x)\) in terms of \(\cos^{-1}(x)\):
\[ \sin^{-1}(x) = \frac{\pi}{2} - \cos^{-1}(x) \]

3. **Substituting for Negative Values**:
Since we are interested in negative values of \(x\), we can express \(-x\) where \(x < 0\). Thus, we can write:
\[ \sin^{-1}(-x) = -\sin^{-1}(x) \]
This is a property of the inverse sine function.

4. **Finding \(\sin^{-1}(x)\) for Negative \(x\)**:
If we let \(x = -y\) where \(y\) is positive, we can write:
\[ \sin^{-1}(-y) = -\sin^{-1}(y) \]

5. **Using the Cosine Identity**:
We also know that:
\[ \sin^{-1}(x) = -\cos^{-1}(\sqrt{1 - x^2}) \]
for negative values of \(x\). Therefore, we can express \(\sin^{-1}(x)\) in terms of \(\cos^{-1}\):
\[ \sin^{-1}(x) = -\cos^{-1}(\sqrt{1 - x^2}) \]

6. **Final Expression**:
Thus, we conclude that:
\[ \sin^{-1}(x) = -\cos^{-1}(\sqrt{1 - x^2}) \]

### Conclusion:
When \(x\) takes negative permissible values, we have:
\[ \sin^{-1}(x) = -\cos^{-1}(\sqrt{1 - x^2}) \]