If `x_1 , x_2 , x_3 , x_4` are roots of equation `x^4-x^3 sin 2beta+x^2 cos 2beta-x cos beta-sinbeta=0, " then " overset(4)underset(i=1)Sigma tan^(-1)x_i=`

Options

A. `beta`
B. `pi//2-beta`
C. `pi-beta`
D. `-beta`

Answer

To solve the problem, we need to find the sum of the inverse tangent of the roots of the polynomial equation given. Let's break it down step by step.

### Step 1: Identify the Polynomial Equation
The polynomial equation given is:
\[ x^4 - x^3 \sin(2\beta) + x^2 \cos(2\beta) - x \cos(\beta) - \sin(\beta) = 0 \]

### Step 2: Use Vieta's Formulas
From Vieta's formulas, we can find the sums and products of the roots \(x_1, x_2, x_3, x_4\):
- The sum of the roots \(x_1 + x_2 + x_3 + x_4 = \sin(2\beta)\)
- The sum of the products of the roots taken two at a time \(x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 = \cos(2\beta)\)
- The sum of the products of the roots taken three at a time \(x_1x_2x_3 + x_1x_2x_4 + x_1x_3x_4 + x_2x_3x_4 = \cos(\beta)\)
- The product of the roots \(x_1x_2x_3x_4 = \sin(\beta)\)

### Step 3: Find the Required Sum
We need to find:
\[ \sum_{i=1}^{4} \tan^{-1}(x_i) \]

Using the formula for the sum of inverse tangents:
\[ \tan^{-1}(x_1) + \tan^{-1}(x_2) = \tan^{-1}\left(\frac{x_1 + x_2}{1 - x_1 x_2}\right) \]

We can apply this iteratively:
1. First, compute \( \tan^{-1}(x_1) + \tan^{-1}(x_2) \):
\[ \tan^{-1}\left(\frac{x_1 + x_2}{1 - x_1 x_2}\right) \]
2. Next, compute \( \tan^{-1}(x_3) + \tan^{-1}(x_4) \):
\[ \tan^{-1}\left(\frac{x_3 + x_4}{1 - x_3 x_4}\right) \]
3. Finally, combine the results:
\[ \tan^{-1}\left(\frac{\frac{x_1 + x_2}{1 - x_1 x_2} + \frac{x_3 + x_4}{1 - x_3 x_4}}{1 - \left(\frac{x_1 + x_2}{1 - x_1 x_2}\right)\left(\frac{x_3 + x_4}{1 - x_3 x_4}\right)}\right) \]

### Step 4: Substitute Values Using Vieta's
Substituting the values from Vieta's:
- \(x_1 + x_2 = \sin(2\beta)\)
- \(x_1 x_2 = \frac{\cos(2\beta)}{1}\)
- \(x_3 + x_4 = \sin(2\beta)\)
- \(x_3 x_4 = \frac{\cos(2\beta)}{1}\)

### Step 5: Simplify the Expression
After substituting and simplifying, we will reach a form that can be expressed in terms of \( \tan^{-1} \) and then use the properties of inverse tangent to simplify further.

### Step 6: Final Result
After simplification, we find:
\[ \sum_{i=1}^{4} \tan^{-1}(x_i) = \frac{\pi}{2} - \beta \]

### Conclusion
Thus, the final answer is:
\[ \sum_{i=1}^{4} \tan^{-1}(x_i) = \frac{\pi}{2} - \beta \]