`sin^(-1) sqrt(2-x)=cos^(-1) sqrt(x-1)` holds for all real x.True or False

Answer

To determine whether the equation \( \sin^{-1}(\sqrt{2 - x}) = \cos^{-1}(\sqrt{x - 1}) \) holds for all real \( x \), we will analyze both sides of the equation step by step.

### Step 1: Define the Variables
Let:
- \( \alpha = \sin^{-1}(\sqrt{2 - x}) \)
- \( \beta = \cos^{-1}(\sqrt{x - 1}) \)

### Step 2: Square Both Sides
From the definition of the inverse functions, we can write:
- \( \sin(\alpha) = \sqrt{2 - x} \)
- \( \cos(\beta) = \sqrt{x - 1} \)

### Step 3: Express \( x \) in Terms of \( \alpha \) and \( \beta \)
Squaring both sides gives us:
1. From \( \sin(\alpha) = \sqrt{2 - x} \):
\[ \sin^2(\alpha) = 2 - x \implies x = 2 - \sin^2(\alpha) \]
2. From \( \cos(\beta) = \sqrt{x - 1} \):
\[ \cos^2(\beta) = x - 1 \implies x = 1 + \cos^2(\beta) \]

### Step 4: Set the Two Expressions for \( x \) Equal
Equating the two expressions for \( x \):
\[ 2 - \sin^2(\alpha) = 1 + \cos^2(\beta) \]

### Step 5: Use the Pythagorean Identity
Using the identity \( \sin^2(\alpha) + \cos^2(\alpha) = 1 \), we can express \( \cos^2(\beta) \) as:
\[ \cos^2(\beta) = 1 - \sin^2(\beta) \]
Thus, we can rewrite the equation:
\[ 2 - \sin^2(\alpha) = 1 + (1 - \sin^2(\beta)) \]
This simplifies to:
\[ 2 - \sin^2(\alpha) = 2 - \sin^2(\beta) \]

### Step 6: Conclude the Relationship
From the above, we have:
\[ \sin^2(\alpha) = \sin^2(\beta) \]
This implies:
\[ \alpha = \beta \quad \text{or} \quad \alpha = \pi - \beta \]

### Step 7: Determine the Validity of the Original Statement
Since \( \alpha = \sin^{-1}(\sqrt{2 - x}) \) and \( \beta = \cos^{-1}(\sqrt{x - 1}) \), we need to check the domains of both sides:
1. For \( \sin^{-1}(\sqrt{2 - x}) \) to be valid, \( \sqrt{2 - x} \) must be in the range \([-1, 1]\), which implies:
\[ 0 \leq 2 - x \leq 1 \implies 1 \leq x \leq 2 \]
2. For \( \cos^{-1}(\sqrt{x - 1}) \) to be valid, \( \sqrt{x - 1} \) must also be in the range \([0, 1]\), leading to:
\[ 0 \leq x - 1 \leq 1 \implies 1 \leq x \leq 2 \]

### Conclusion
Both expressions are equal only for \( x \) in the interval \([1, 2]\). Therefore, the statement \( \sin^{-1}(\sqrt{2 - x}) = \cos^{-1}(\sqrt{x - 1}) \) does not hold for all real \( x \).

Thus, the answer is **False**.

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