If `sin^4x + cos^4y + 2 = 4 sin x.cos y` and `0 le x, y le pi/2` then sin x + cos y is equal to

Options

A. `-2`
B. 0
C. 2
D. none of these

Explanation

To solve the equation \( \sin^4 x + \cos^4 y + 2 = 4 \sin x \cos y \) for \( \sin x + \cos y \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \sin^4 x + \cos^4 y + 2 = 4 \sin x \cos y \]
Notice that we can express \( \sin^4 x \) and \( \cos^4 y \) in a different form:
\[ \sin^4 x = (\sin^2 x)^2 \quad \text{and} \quad \cos^4 y = (\cos^2 y)^2 \]
Thus, we can rewrite the equation as:
\[ (\sin^2 x)^2 + (\cos^2 y)^2 + 2 = 4 \sin x \cos y \]

### Step 2: Introduce substitutions
Let \( a = \sin^2 x \) and \( b = \cos^2 y \). The equation becomes:
\[ a^2 + b^2 + 2 = 4 \sqrt{a} \sqrt{b} \]

### Step 3: Rearrange the equation
Rearranging gives us:
\[ a^2 + b^2 - 4 \sqrt{ab} + 2 = 0 \]

### Step 4: Recognize a perfect square
Notice that we can express \( a^2 - 4\sqrt{ab} + b^2 \) as a perfect square:
\[ (a - 2\sqrt{b})^2 + 2 = 0 \]
This implies:
\[ (a - 2\sqrt{b})^2 = -2 \]
Since the square of a real number cannot be negative, we need to consider the conditions under which this equation holds.

### Step 5: Set conditions for equality
For the original equation to hold, we must have:
\[ a - 2\sqrt{b} = 0 \quad \text{and} \quad 2 = 0 \]
This leads us to:
\[ \sin^2 x = 2 \cos y \]

### Step 6: Solve for \( \sin x \) and \( \cos y \)
From \( \sin^2 x = 2 \cos^2 y \), we can also express:
\[ \sin^2 x + \cos^2 y = 1 \]
Substituting \( \sin^2 x \) into this equation gives:
\[ 2 \cos^2 y + \cos^2 y = 1 \implies 3 \cos^2 y = 1 \implies \cos^2 y = \frac{1}{3} \]
Thus, \( \cos y = \frac{1}{\sqrt{3}} \).

### Step 7: Find \( \sin x \)
Using \( \sin^2 x = 2 \cos^2 y \):
\[ \sin^2 x = 2 \cdot \frac{1}{3} = \frac{2}{3} \implies \sin x = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3} \]

### Step 8: Calculate \( \sin x + \cos y \)
Now we can find:
\[ \sin x + \cos y = \frac{\sqrt{6}}{3} + \frac{1}{\sqrt{3}} = \frac{\sqrt{6}}{3} + \frac{\sqrt{3}}{3} = \frac{\sqrt{6} + \sqrt{3}}{3} \]

### Conclusion
Thus, the value of \( \sin x + \cos y \) is:
\[ \sin x + \cos y = \frac{\sqrt{6} + \sqrt{3}}{3} \]