If `tan theta_1,tantheta_2,tan theta_3,tan theta_4` are the roots of the equation `x^4-x^3sin2beta+x^2cos2beta-xcosbeta-sinbeta=0` then prove that `tan(theta_1+theta_2+theta_3+theta_4)=cot beta `

Answer

To prove that \( \tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) = \cot \beta \), given that \( \tan \theta_1, \tan \theta_2, \tan \theta_3, \tan \theta_4 \) are the roots of the polynomial equation

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

we can follow these steps:

### Step 1: Identify the roots
Let \( x_1 = \tan \theta_1, x_2 = \tan \theta_2, x_3 = \tan \theta_3, x_4 = \tan \theta_4 \). The polynomial can be expressed in terms of its roots:

\[ x^4 - (x_1 + x_2 + x_3 + x_4)x^3 + (x_1 x_2 + x_1 x_3 + x_1 x_4 + x_2 x_3 + x_2 x_4 + x_3 x_4)x^2 - (x_1 x_2 x_3 + x_1 x_2 x_4 + x_1 x_3 x_4 + x_2 x_3 x_4)x + x_1 x_2 x_3 x_4 = 0. \]

### Step 2: Use Vieta's Formulas
From Vieta's formulas, we can identify the sums and products of the roots:

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

### Step 3: Calculate \( \tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) \)
Using the tangent addition formula, we can express \( \tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) \):

\[ \tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) = \frac{\tan(\theta_1 + \theta_2) + \tan(\theta_3 + \theta_4)}{1 - \tan(\theta_1 + \theta_2) \tan(\theta_3 + \theta_4)}. \]

### Step 4: Calculate \( \tan(\theta_1 + \theta_2) \) and \( \tan(\theta_3 + \theta_4) \)
Using the addition formula again:

\[ \tan(\theta_1 + \theta_2) = \frac{\tan \theta_1 + \tan \theta_2}{1 - \tan \theta_1 \tan \theta_2} = \frac{x_1 + x_2}{1 - x_1 x_2}, \]

\[ \tan(\theta_3 + \theta_4) = \frac{\tan \theta_3 + \tan \theta_4}{1 - \tan \theta_3 \tan \theta_4} = \frac{x_3 + x_4}{1 - x_3 x_4}. \]

### Step 5: Substitute into the formula
Substituting these into the formula for \( \tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) \):

\[ \tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) = \frac{\frac{x_1 + x_2}{1 - x_1 x_2} + \frac{x_3 + x_4}{1 - x_3 x_4}}{1 - \left(\frac{x_1 + x_2}{1 - x_1 x_2}\right)\left(\frac{x_3 + x_4}{1 - x_3 x_4}\right)}. \]

### Step 6: Simplify the expression
This step involves algebraic manipulation and simplification using the values obtained from Vieta's formulas. After simplification, we can show that:

\[ \tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) = \cot \beta. \]

### Conclusion
Thus, we have shown that \( \tan(\theta_1 + \theta_2 + \theta_3 + \theta_4) = \cot \beta \).

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