Let `0 lt=thetalt=pi/2 and x=X cos theta+Y sin theta, y=X sin theta-Y cos theta` such that `x^2+2xy+y^2=aX^2+bY^2,` where a and b are constant, then

Options

A. `a=-1, b=-3`
B. `theta=pi/2`
C. `a=2, b=0`
D. `theta=pi/3`

Answer

To solve the problem, we start with the given equations and manipulate them step by step.

### Step 1: Write the given equations
We have:
- \( x = X \cos \theta + Y \sin \theta \)
- \( y = X \sin \theta - Y \cos \theta \)

We need to analyze the expression:
\[ x^2 + 2xy + y^2 = aX^2 + bY^2 \]

### Step 2: Substitute \( x \) and \( y \) into the equation
Substituting the values of \( x \) and \( y \) into the left-hand side:
\[ x^2 + 2xy + y^2 = (X \cos \theta + Y \sin \theta)^2 + 2(X \cos \theta + Y \sin \theta)(X \sin \theta - Y \cos \theta) + (X \sin \theta - Y \cos \theta)^2 \]

### Step 3: Expand the terms
Now we expand each term:
1. \( (X \cos \theta + Y \sin \theta)^2 = X^2 \cos^2 \theta + 2XY \cos \theta \sin \theta + Y^2 \sin^2 \theta \)
2. \( (X \sin \theta - Y \cos \theta)^2 = X^2 \sin^2 \theta - 2XY \sin \theta \cos \theta + Y^2 \cos^2 \theta \)
3. \( 2(X \cos \theta + Y \sin \theta)(X \sin \theta - Y \cos \theta) = 2(X^2 \cos \theta \sin \theta - Y^2 \sin \theta \cos \theta + XY (\sin^2 \theta - \cos^2 \theta)) \)

### Step 4: Combine all the expanded terms
Combining all the terms gives:
\[ x^2 + 2xy + y^2 = (X^2 \cos^2 \theta + Y^2 \sin^2 \theta + 2XY \cos \theta \sin \theta) + (X^2 \sin^2 \theta + Y^2 \cos^2 \theta - 2XY \sin \theta \cos \theta) + 2(X^2 \cos \theta \sin \theta - Y^2 \sin \theta \cos \theta + XY (\sin^2 \theta - \cos^2 \theta)) \]

### Step 5: Simplify the expression
Now we simplify:
\[ = X^2 (\cos^2 \theta + \sin^2 \theta) + Y^2 (\sin^2 \theta + \cos^2 \theta) + 2XY (\cos \theta \sin \theta - \sin \theta \cos \theta + \sin^2 \theta - \cos^2 \theta) \]
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ = X^2 + Y^2 + 2XY (\sin^2 \theta - \cos^2 \theta) \]

### Step 6: Set the equation equal to the right-hand side
Now we have:
\[ X^2 + Y^2 + 2XY (\sin^2 \theta - \cos^2 \theta) = aX^2 + bY^2 \]

### Step 7: Compare coefficients
From here, we can compare coefficients:
1. Coefficient of \( X^2 \): \( 1 = a \)
2. Coefficient of \( Y^2 \): \( 1 = b \)
3. Coefficient of \( XY \): \( 2(\sin^2 \theta - \cos^2 \theta) = 0 \)

### Step 8: Solve for \( \theta \)
The equation \( 2(\sin^2 \theta - \cos^2 \theta) = 0 \) implies:
\[ \sin^2 \theta = \cos^2 \theta \implies \tan^2 \theta = 1 \implies \theta = \frac{\pi}{4} \]

### Step 9: Determine constants \( a \) and \( b \)
From the comparisons:
- \( a = 1 \)
- \( b = 1 \)

### Final Result
Thus, we conclude:
- \( a = 1 \)
- \( b = 1 \)
- \( \theta = \frac{\pi}{4} \)