If `sin^(-1)x+sin^(-1)(1-x)=cos^(-1)x ` then x equals

Options

A. 1,-1
B. 1,0
C. `0,1/2`
D. none of these

Answer

To solve the equation \( \sin^{-1} x + \sin^{-1} (1 - x) = \cos^{-1} x \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \sin^{-1} x + \sin^{-1} (1 - x) = \cos^{-1} x \]

### Step 2: Use the identity for sine and cosine
We know that:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
Using this identity, we can rewrite the left-hand side:
\[ \sin^{-1} x + \sin^{-1} (1 - x) = \frac{\pi}{2} - \sin^{-1} (1 - x) \]
Thus, we can rewrite the equation as:
\[ \frac{\pi}{2} - \sin^{-1} (1 - x) = \cos^{-1} x \]

### Step 3: Rearranging the equation
Rearranging gives us:
\[ \sin^{-1} (1 - x) = \frac{\pi}{2} - \cos^{-1} x \]
Using the identity again, we can simplify this to:
\[ \sin^{-1} (1 - x) = \sin^{-1} x \]

### Step 4: Equate the arguments of sine
Since the sine function is one-to-one in the range of \([-1, 1]\), we can equate the arguments:
\[ 1 - x = x \]

### Step 5: Solve for \(x\)
Solving the equation \(1 - x = x\):
\[ 1 = 2x \implies x = \frac{1}{2} \]

### Step 6: Check for other solutions
We also need to check if there are any other solutions. Since both \( \sin^{-1} x \) and \( \sin^{-1} (1 - x) \) are defined for \( x \in [0, 1] \), we can check the endpoints:
- If \( x = 0 \):
\[ \sin^{-1}(0) + \sin^{-1}(1) = 0 + \frac{\pi}{2} = \frac{\pi}{2} \quad \text{and} \quad \cos^{-1}(0) = \frac{\pi}{2} \]
This holds true.

- If \( x = 1 \):
\[ \sin^{-1}(1) + \sin^{-1}(0) = \frac{\pi}{2} + 0 = \frac{\pi}{2} \quad \text{and} \quad \cos^{-1}(1) = 0 \]
This does not hold.

### Conclusion
The solutions to the equation are:
\[ x = 0 \quad \text{and} \quad x = \frac{1}{2} \]