If `n` be a natural number define polynomial `f_(n)(x)` of `n^(th)` degree as follows `f_(n)(costheta)cosntheta`
i.e. `f_(2)(x)=2x^(2)-1 f_(3)(x)=4x^(3)-3x_(1)`
Then
`(x+sqrt(x^(2)-1))^(10)+(x-sqrt(x^(2)-1s))^(10)` is equal to

Options

A. `f_(10)(x)`
B. `f_(11)(x)+f_(9)(x)`
C. `f_(11)(x)`
D. `2f_(10)(x)`

Explanation

To solve the problem, we need to evaluate the expression \((x + \sqrt{x^2 - 1})^{10} + (x - \sqrt{x^2 - 1})^{10}\).

### Step-by-step Solution:

1. **Recognize the Structure**: Notice that the expression can be rewritten in terms of trigonometric functions. Let \( x = \cos \theta \). Then, we have:
\[ \sqrt{x^2 - 1} = \sqrt{\cos^2 \theta - 1} = \sqrt{-\sin^2 \theta} = i \sin \theta \]
Thus, we can rewrite the expression as:
\[ (x + i \sin \theta)^{10} + (x - i \sin \theta)^{10} \]

2. **Use Euler's Formula**: We can express \( x + i \sin \theta \) and \( x - i \sin \theta \) using Euler's formula:
\[ x + i \sin \theta = \cos \theta + i \sin \theta = e^{i\theta} \]
\[ x - i \sin \theta = \cos \theta - i \sin \theta = e^{-i\theta} \]

3. **Raise to the Power of 10**: Now, we can raise both expressions to the power of 10:
\[ (e^{i\theta})^{10} + (e^{-i\theta})^{10} = e^{10i\theta} + e^{-10i\theta} \]

4. **Simplify Using Cosine**: The sum of exponentials can be simplified using the cosine function:
\[ e^{10i\theta} + e^{-10i\theta} = 2 \cos(10\theta) \]

5. **Relate Back to the Polynomial**: We know that the polynomial \( f_n(x) \) is defined in terms of cosine. Specifically, we have:
\[ f_{10}(x) = 2 \cos(10\theta) \]
Therefore, the expression simplifies to:
\[ (x + \sqrt{x^2 - 1})^{10} + (x - \sqrt{x^2 - 1})^{10} = 2 f_{10}(x) \]

### Final Result:
Thus, the final answer is:
\[ 2 f_{10}(x) \]