`cos^(- 1)sqrt((a-x)/(a-b))=sin^(- 1)sqrt((x-b)/(a-b))` is possible ,if

Options

A. `a gt x gt b`
B. `a lt x lt b`
C. `a = x = b`
D. `a gt b` and x takes any value

Explanation

To solve the equation \( \cos^{-1} \sqrt{\frac{a-x}{a-b}} = \sin^{-1} \sqrt{\frac{x-b}{a-b}} \), we need to analyze the conditions under which this equality holds true.

### Step-by-Step Solution:

1. **Understanding the Functions**:
The equation involves the inverse cosine and inverse sine functions. We know that:
\[ \cos^{-1}(y) + \sin^{-1}(y) = \frac{\pi}{2} \]
for \( y \) in the range \([0, 1]\). This means that if we set:
\[ y = \sqrt{\frac{x-b}{a-b}} \]
then:
\[ \cos^{-1} \sqrt{\frac{a-x}{a-b}} = \frac{\pi}{2} - \sin^{-1} \sqrt{\frac{x-b}{a-b}} \]

2. **Setting Up the Inequalities**:
For the inverse trigonometric functions to be defined, we need:
\[ 0 \leq \sqrt{\frac{a-x}{a-b}} \leq 1 \]
and
\[ 0 \leq \sqrt{\frac{x-b}{a-b}} \leq 1 \]

3. **Analyzing the First Inequality**:
From \( 0 \leq \sqrt{\frac{a-x}{a-b}} \leq 1 \):
- The left side gives us \( a-x \geq 0 \) or \( a \geq x \).
- The right side gives us \( a-x \leq a-b \) or \( x \leq b \).
Thus, we have:
\[ b \leq x \leq a \]

4. **Analyzing the Second Inequality**:
From \( 0 \leq \sqrt{\frac{x-b}{a-b}} \leq 1 \):
- The left side gives us \( x-b \geq 0 \) or \( x \geq b \).
- The right side gives us \( x-b \leq a-b \) or \( x \leq a \).
Thus, we have:
\[ b \leq x \leq a \]

5. **Combining the Results**:
From both inequalities, we conclude that:
\[ b \leq x \leq a \]

### Conclusion:
The equation \( \cos^{-1} \sqrt{\frac{a-x}{a-b}} = \sin^{-1} \sqrt{\frac{x-b}{a-b}} \) is possible if \( b \leq x \leq a \).