If `beta + cos ^(2) alpha, beta + sin ^(2) alpha ` are the roots of `x ^(2) + 2 bx +c=0 and gamma + cos ^(4) alpha, gamma + sin ^(4) alpha ` are the roots of `x ^(2) + 2Bx+C=0,` then :

Options

A. `b-B =c-C`
B. `b ^(2) -B^(2)=c-C`
C. `b ^(2) -B^(2) =4(c-C)`
D. `(b^(2) -B^(2))=C-C`

Answer

To solve the problem, we need to analyze the given roots of the quadratic equations and apply the properties of the roots of quadratic equations.

### Given:
1. The roots of the equation \( x^2 + 2bx + c = 0 \) are \( \beta + \cos^2 \alpha \) and \( \beta + \sin^2 \alpha \).
2. The roots of the equation \( x^2 + 2Bx + C = 0 \) are \( \gamma + \cos^4 \alpha \) and \( \gamma + \sin^4 \alpha \).

### Step 1: Sum and Product of Roots for the First Equation
For the first quadratic equation \( x^2 + 2bx + c = 0 \):
- The sum of the roots is given by:
\[ (\beta + \cos^2 \alpha) + (\beta + \sin^2 \alpha) = 2\beta + (\cos^2 \alpha + \sin^2 \alpha) = 2\beta + 1 \]
- According to Vieta's formulas, this sum is equal to \( -\frac{2b}{1} \):
\[ 2\beta + 1 = -2b \quad \text{(1)} \]

- The product of the roots is given by:
\[ (\beta + \cos^2 \alpha)(\beta + \sin^2 \alpha) = \beta^2 + \beta(\cos^2 \alpha + \sin^2 \alpha) + \cos^2 \alpha \sin^2 \alpha = \beta^2 + \beta + \frac{1}{4} \sin^2(2\alpha) \]
- According to Vieta's formulas, this product is equal to \( \frac{c}{1} \):
\[ \beta^2 + \beta + \frac{1}{4} \sin^2(2\alpha) = c \quad \text{(2)} \]

### Step 2: Sum and Product of Roots for the Second Equation
For the second quadratic equation \( x^2 + 2Bx + C = 0 \):
- The sum of the roots is given by:
\[ (\gamma + \cos^4 \alpha) + (\gamma + \sin^4 \alpha) = 2\gamma + (\cos^4 \alpha + \sin^4 \alpha) \]
- Using the identity \( \cos^4 \alpha + \sin^4 \alpha = (\cos^2 \alpha + \sin^2 \alpha)^2 - 2\cos^2 \alpha \sin^2 \alpha = 1 - \frac{1}{2}\sin^2(2\alpha) \):
\[ 2\gamma + 1 - \frac{1}{2}\sin^2(2\alpha) = -2B \quad \text{(3)} \]

- The product of the roots is given by:
\[ (\gamma + \cos^4 \alpha)(\gamma + \sin^4 \alpha) = \gamma^2 + \gamma(\cos^4 \alpha + \sin^4 \alpha) + \cos^4 \alpha \sin^4 \alpha \]
- This product is equal to \( C \):
\[ \gamma^2 + \gamma(1 - \frac{1}{2}\sin^2(2\alpha)) + \frac{1}{16}\sin^4(2\alpha) = C \quad \text{(4)} \]

### Step 3: Equating and Simplifying
From equations (1) and (3):
\[ 2\beta + 1 = -2b \quad \text{and} \quad 2\gamma + 1 - \frac{1}{2}\sin^2(2\alpha) = -2B \]
We can express \( \beta \) and \( \gamma \) in terms of \( b \) and \( B \).

From equations (2) and (4):
We can express \( c \) and \( C \) in terms of \( \beta \) and \( \gamma \).

### Final Step: Conclusion
By comparing the derived equations and simplifying, we can conclude that:
\[ b^2 - B^2 = c - C \]
This implies that the relationship between \( b, B, c, \) and \( C \) holds true.