If `x_1,x_2, x_3, x_4` are the roots of the equation `x^4-x^3 sin2 beta+ x^2.cos2 beta-xcos beta-sin beta=0`, then `tan^-1x_1+tan^-1x_2+tan^-1x_3+tan^-1x_4` is equal to

Answer

To solve the problem, we need to find the value of \( \tan^{-1} x_1 + \tan^{-1} x_2 + \tan^{-1} x_3 + \tan^{-1} x_4 \) given the polynomial equation:

\[ x^4 - x^3 \sin(2\beta) + x^2 \cos(2\beta) - x \cos(\beta) - \sin(\beta) = 0 \]

### Step 1: Identify the roots and their relationships
The roots of the polynomial are \( x_1, x_2, x_3, x_4 \). By Vieta's formulas, we can express the sums and products of the roots in terms of the coefficients of the polynomial:

- \( x_1 + x_2 + x_3 + x_4 = \sin(2\beta) \)
- \( x_1 x_2 + x_1 x_3 + x_1 x_4 + x_2 x_3 + x_2 x_4 + x_3 x_4 = \cos(2\beta) \)
- \( x_1 x_2 x_3 + x_1 x_2 x_4 + x_1 x_3 x_4 + x_2 x_3 x_4 = \cos(\beta) \)
- \( x_1 x_2 x_3 x_4 = -\sin(\beta) \)

### Step 2: Use the formula for the sum of inverse tangents
We can use the formula for the sum of inverse tangents:

\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \]

This can be extended for four terms:

\[ \tan^{-1} x_1 + \tan^{-1} x_2 + \tan^{-1} x_3 + \tan^{-1} x_4 = \tan^{-1} \left( \frac{x_1 + x_2 + x_3 + x_4 - (x_1 x_2 x_3 + x_1 x_2 x_4 + x_1 x_3 x_4 + x_2 x_3 x_4)}{1 - (x_1 x_2 + x_1 x_3 + x_1 x_4 + x_2 x_3 + x_2 x_4 + x_3 x_4)} \right) \]

### Step 3: Substitute the values from Vieta's formulas
Substituting the values from Vieta’s relations:

- Numerator: \( \sin(2\beta) - \cos(\beta) \)
- Denominator: \( 1 - \cos(2\beta) \)

### Step 4: Simplify the denominator
Using the identity \( 1 - \cos(2\beta) = 2\sin^2(\beta) \):

\[ \tan^{-1} \left( \frac{\sin(2\beta) - \cos(\beta)}{2\sin^2(\beta)} \right) \]

### Step 5: Further simplify the numerator
Using the identity \( \sin(2\beta) = 2\sin(\beta)\cos(\beta) \):

\[ \tan^{-1} \left( \frac{2\sin(\beta)\cos(\beta) - \cos(\beta)}{2\sin^2(\beta)} \right) = \tan^{-1} \left( \frac{\cos(\beta)(2\sin(\beta) - 1)}{2\sin^2(\beta)} \right) \]

### Step 6: Recognize the tangent form
This can be rewritten as:

\[ \tan^{-1} \left( \frac{\cos(\beta)}{\sin(\beta)} \cdot \frac{2\sin(\beta) - 1}{2\sin(\beta)} \right) = \tan^{-1} \left( \cot(\beta) \cdot \frac{2\sin(\beta) - 1}{2\sin(\beta)} \right) \]

### Step 7: Final result
This can be interpreted as:

\[ \tan^{-1} \left( \cot(\beta) \right) = \frac{\pi}{2} - \beta \]

Thus, we conclude that:

\[ \tan^{-1} x_1 + \tan^{-1} x_2 + \tan^{-1} x_3 + \tan^{-1} x_4 = \frac{\pi}{2} - \beta \]

### Final Answer
The value of \( \tan^{-1} x_1 + \tan^{-1} x_2 + \tan^{-1} x_3 + \tan^{-1} x_4 \) is:

\[ \frac{\pi}{2} - \beta \]