If `xepsilon(0,1)` and `f(x)=sec{tan^(-1)((sin(cos^(-1)x)+cos(sin^(-1)x))/(cos(cos^(-1)x)=sin(sin^(-1)x)))}`, then `sum_(r=2)^(10)f(1/r)` is

Explanation

To solve the problem, we need to simplify the function \( f(x) \) given by:

\[ f(x) = \sec\left(\tan^{-1}\left(\frac{\sin(\cos^{-1}x) + \cos(\sin^{-1}x)}{\cos(\cos^{-1}x) + \sin(\sin^{-1}x)}\right)\right) \]

### Step 1: Simplifying the terms inside the function

1. **Identify the terms**:
- \( \sin(\cos^{-1}x) = \sqrt{1 - x^2} \)
- \( \cos(\sin^{-1}x) = \sqrt{1 - x^2} \)
- \( \cos(\cos^{-1}x) = x \)
- \( \sin(\sin^{-1}x) = x \)

2. **Substituting these values**:
\[ f(x) = \sec\left(\tan^{-1}\left(\frac{\sqrt{1 - x^2} + \sqrt{1 - x^2}}{x + x}\right)\right) \]
\[ = \sec\left(\tan^{-1}\left(\frac{2\sqrt{1 - x^2}}{2x}\right)\right) \]
\[ = \sec\left(\tan^{-1}\left(\frac{\sqrt{1 - x^2}}{x}\right)\right) \]

### Step 2: Using the identity for secant

3. **Using the identity**:
- If \( \theta = \tan^{-1}\left(\frac{\sqrt{1 - x^2}}{x}\right) \), then \( \sec(\theta) = \frac{1}{\cos(\theta)} \).
- In a right triangle where the opposite side is \( \sqrt{1 - x^2} \) and the adjacent side is \( x \), the hypotenuse is \( \sqrt{(\sqrt{1 - x^2})^2 + x^2} = 1 \).

4. **Thus, we have**:
\[ f(x) = \frac{1}{x} \]

### Step 3: Finding the summation

5. **Now substituting \( x = \frac{1}{r} \)**:
\[ f\left(\frac{1}{r}\right) = r \]

6. **Calculating the summation from \( r = 2 \) to \( r = 10 \)**:
\[ \sum_{r=2}^{10} f\left(\frac{1}{r}\right) = \sum_{r=2}^{10} r = 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 \]

### Step 4: Performing the summation

7. **Calculating the sum**:
\[ = 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 54 \]

### Final Answer

Thus, the final answer is:

\[ \sum_{r=2}^{10} f\left(\frac{1}{r}\right) = 54 \]