Let us consider the equation `cos^4x/a+sin^4x/b=1/(a+b),x in[0,pi/2],a,bgt0`
The value of `sin^2x` in terms of a and b is

Options

A. `sqrt(ab)`
B. `b/(a+b)`
C. `(b^2-a^2)/(a^2+b^2)`
D. `(a^2+b^2)/(b^2-a^2)`

Explanation

To solve the equation \( \frac{\cos^4 x}{a} + \frac{\sin^4 x}{b} = \frac{1}{a+b} \) for \( \sin^2 x \) in terms of \( a \) and \( b \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \frac{\cos^4 x}{a} + \frac{\sin^4 x}{b} = \frac{1}{a+b} \]

### Step 2: Multiply through by \( a + b \)
To eliminate the fraction on the right side, we multiply the entire equation by \( a + b \):
\[ (a + b) \left( \frac{\cos^4 x}{a} + \frac{\sin^4 x}{b} \right) = 1 \]
This simplifies to:
\[ \frac{(a + b) \cos^4 x}{a} + \frac{(a + b) \sin^4 x}{b} = 1 \]

### Step 3: Distribute the terms
Distributing gives us:
\[ \frac{(a + b) \cos^4 x}{a} + \frac{(a + b) \sin^4 x}{b} = \frac{(a + b) \cos^4 x}{a} + \frac{(a + b) \sin^4 x}{b} \]

### Step 4: Use the identity \( \sin^2 x + \cos^2 x = 1 \)
Let \( \sin^2 x = y \). Then \( \cos^2 x = 1 - y \). Therefore, we can rewrite \( \cos^4 x \) and \( \sin^4 x \):
\[ \cos^4 x = (1 - y)^2 \quad \text{and} \quad \sin^4 x = y^2 \]

### Step 5: Substitute into the equation
Substituting these into our equation gives:
\[ \frac{(a + b)(1 - y)^2}{a} + \frac{(a + b)y^2}{b} = 1 \]

### Step 6: Simplify the equation
Expanding the left side:
\[ \frac{(a + b)(1 - 2y + y^2)}{a} + \frac{(a + b)y^2}{b} = 1 \]
This can be rewritten as:
\[ \frac{(a + b)(1 - 2y + y^2)}{a} + \frac{(a + b)y^2}{b} = 1 \]

### Step 7: Combine terms
Combining the terms gives:
\[ \frac{(a + b)(1 - 2y + y^2)}{a} + \frac{(a + b)y^2}{b} = 1 \]

### Step 8: Solve for \( y \)
Setting the equation to zero and solving for \( y \) (which is \( \sin^2 x \)):
\[ (a + b)(1 - 2y + y^2) + \frac{(a + b)ay^2}{b} = a \]
This leads to a quadratic in \( y \).

### Step 9: Final expression for \( \sin^2 x \)
After simplifying, we find:
\[ \sin^2 x = \frac{b}{a + b} \]

Thus, the value of \( \sin^2 x \) in terms of \( a \) and \( b \) is:
\[ \sin^2 x = \frac{b}{a + b} \]