If `x=acostheta+bsinthetaandy=bcostheta-asintheta`,then `x^(2)+y^(2)` is equal to

Options

A. ab
B. `a^(2)+b^(2)`
C. `a^(2)-b^(2)`
D. 1

Answer

To solve the problem, we need to find the value of \( x^2 + y^2 \) given the equations:

\[ x = a \cos \theta + b \sin \theta \]
\[ y = b \cos \theta - a \sin \theta \]

### Step 1: Square both equations

First, we will square both \( x \) and \( y \):

\[ x^2 = (a \cos \theta + b \sin \theta)^2 \]
\[ y^2 = (b \cos \theta - a \sin \theta)^2 \]

### Step 2: Expand the squares

Now, we will expand both squares:

\[ x^2 = a^2 \cos^2 \theta + 2ab \cos \theta \sin \theta + b^2 \sin^2 \theta \]
\[ y^2 = b^2 \cos^2 \theta - 2ab \cos \theta \sin \theta + a^2 \sin^2 \theta \]

### Step 3: Add the two equations

Now, we will add \( x^2 \) and \( y^2 \):

\[ x^2 + y^2 = (a^2 \cos^2 \theta + 2ab \cos \theta \sin \theta + b^2 \sin^2 \theta) + (b^2 \cos^2 \theta - 2ab \cos \theta \sin \theta + a^2 \sin^2 \theta) \]

### Step 4: Combine like terms

Now, we will combine the like terms:

\[ x^2 + y^2 = a^2 \cos^2 \theta + b^2 \cos^2 \theta + b^2 \sin^2 \theta + a^2 \sin^2 \theta + (2ab \cos \theta \sin \theta - 2ab \cos \theta \sin \theta) \]

Notice that the \( 2ab \cos \theta \sin \theta \) terms cancel out:

\[ x^2 + y^2 = (a^2 \cos^2 \theta + a^2 \sin^2 \theta) + (b^2 \cos^2 \theta + b^2 \sin^2 \theta) \]

### Step 5: Factor out common terms

Using the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \):

\[ x^2 + y^2 = a^2 (\cos^2 \theta + \sin^2 \theta) + b^2 (\cos^2 \theta + \sin^2 \theta) \]
\[ x^2 + y^2 = a^2 \cdot 1 + b^2 \cdot 1 \]
\[ x^2 + y^2 = a^2 + b^2 \]

### Final Answer

Thus, the value of \( x^2 + y^2 \) is:

\[ \boxed{a^2 + b^2} \]