If ` alpha , beta ` be the roots of the the equation ` x^(2) - 2 x + 2 = 0 ` and ` cot theta = x + 1 " then " (( x + alpha )^(n) - ( x + beta)^(n))/( alpha - beta) =`

Options

A. `(cos n theta)/( sin^(n) theta)`
B. `(sin n theta)/(cos ^(n) theta)`
C. `(sin n theta)/(sin^(n)theta)`
D. `(cos n theta)/(cos ^(n) theta)`

Answer

To solve the problem step by step, we will follow the instructions given in the video transcript and derive the required expression.

### Step 1: Find the roots of the quadratic equation
The given quadratic equation is:
\[ x^2 - 2x + 2 = 0 \]

Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -2, c = 2 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-2)^2 - 4 \cdot 1 \cdot 2 = 4 - 8 = -4 \]

Now substituting into the quadratic formula:
\[ x = \frac{2 \pm \sqrt{-4}}{2 \cdot 1} = \frac{2 \pm 2i}{2} = 1 \pm i \]

Thus, the roots are:
\[ \alpha = 1 + i \quad \text{and} \quad \beta = 1 - i \]

### Step 2: Express \( x \) in terms of \( \theta \)
Given that:
\[ \cot \theta = x + 1 \]

We can rearrange this to find \( x \):
\[ x = \cot \theta - 1 \]

### Step 3: Calculate \( x + \alpha \) and \( x + \beta \)
Substituting \( x \) into \( x + \alpha \):
\[ x + \alpha = (\cot \theta - 1) + (1 + i) = \cot \theta + i \]

Now for \( x + \beta \):
\[ x + \beta = (\cot \theta - 1) + (1 - i) = \cot \theta - i \]

### Step 4: Raise to the power of \( n \)
Now we need to raise both expressions to the power of \( n \):
\[ (x + \alpha)^n = (\cot \theta + i)^n \]
\[ (x + \beta)^n = (\cot \theta - i)^n \]

### Step 5: Substitute into the expression
We need to evaluate:
\[ \frac{(x + \alpha)^n - (x + \beta)^n}{\alpha - \beta} \]

First, calculate \( \alpha - \beta \):
\[ \alpha - \beta = (1 + i) - (1 - i) = 2i \]

Now substituting back:
\[ \frac{(\cot \theta + i)^n - (\cot \theta - i)^n}{2i} \]

### Step 6: Simplify the expression
Using the identity for the difference of powers:
\[ a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + \ldots + b^{n-1}) \]
Let \( a = \cot \theta + i \) and \( b = \cot \theta - i \):
\[ a - b = 2i \]

Thus:
\[ \frac{(a - b)(a^{n-1} + a^{n-2}b + \ldots + b^{n-1})}{2i} = a^{n-1} + a^{n-2}b + \ldots + b^{n-1} \]

### Step 7: Final Result
The final result is:
\[ \frac{(\cot \theta + i)^n - (\cot \theta - i)^n}{2i} = \frac{2i \sin(n\theta)}{2i} = \sin(n\theta) \]

### Conclusion
Thus, the final answer is:
\[ \sin(n\theta) \]