`cos^(-1)sqrt((a-x)/(a-b))=sin^(-1)sqrt((x-b)/(a-b))` for all values of x.

Answer

To solve the equation \( \cos^{-1}\left(\sqrt{\frac{a-x}{a-b}}\right) = \sin^{-1}\left(\sqrt{\frac{x-b}{a-b}}\right) \) for all values of \( x \), we will follow these steps:

### Step 1: Understand the Relationship Between Inverse Functions
Recall the identity that relates the inverse cosine and inverse sine functions:
\[ \cos^{-1}(x) = \sin^{-1}(\sqrt{1 - x^2}) \]
This means we can express the left-hand side (LHS) in terms of the right-hand side (RHS).

### Step 2: Rewrite the LHS Using the Identity
Using the identity, we can rewrite the LHS:
\[ \cos^{-1}\left(\sqrt{\frac{a-x}{a-b}}\right) = \sin^{-1}\left(\sqrt{1 - \left(\sqrt{\frac{a-x}{a-b}}\right)^2}\right) \]

### Step 3: Simplify the Expression Inside the Sine Inverse
Now simplify the expression inside the sine inverse:
\[ 1 - \left(\sqrt{\frac{a-x}{a-b}}\right)^2 = 1 - \frac{a-x}{a-b} = \frac{(a-b) - (a-x)}{a-b} = \frac{x - b}{a-b} \]
Thus, we have:
\[ \cos^{-1}\left(\sqrt{\frac{a-x}{a-b}}\right) = \sin^{-1}\left(\sqrt{\frac{x-b}{a-b}}\right) \]

### Step 4: Equate the Two Sides
Now we can equate the two sides:
\[ \sin^{-1}\left(\sqrt{\frac{x-b}{a-b}}\right) = \sin^{-1}\left(\sqrt{\frac{x-b}{a-b}}\right) \]
This shows that both sides are equal.

### Step 5: Check the Validity of the Equation
For the equation to hold for all \( x \), we need to ensure that both sides are defined. The expressions under the square roots must be non-negative:
1. \( \frac{a-x}{a-b} \geq 0 \) implies \( a-x \geq 0 \) or \( x \leq a \).
2. \( \frac{x-b}{a-b} \geq 0 \) implies \( x-b \geq 0 \) or \( x \geq b \).

### Step 6: Determine the Valid Range for \( x \)
From the inequalities:
- \( b \leq x \leq a \)
This means the equation is valid only for \( x \) in the interval \([b, a]\).

### Conclusion
The original statement \( \cos^{-1}\left(\sqrt{\frac{a-x}{a-b}}\right) = \sin^{-1}\left(\sqrt{\frac{x-b}{a-b}}\right) \) is not true for all \( x \in \mathbb{R} \), but it is true for \( x \) in the interval \([b, a]\).