If `x=a cos theta + b sin theta and y=a sin theta - b cos theta.` then` a ^(2) +b^(2) ` is equal to

Options

A. `x ^(2) -y^(2)`
B. `x ^(2) +y^(2)`
C. `(x+y)^(2)`
D. `(x-y)^(2)`

Answer

To solve the problem, we start with the given equations:

1. \( x = a \cos \theta + b \sin \theta \)
2. \( y = a \sin \theta - b \cos \theta \)

We need to find the value of \( a^2 + b^2 \).

### Step 1: Square both equations

First, we square both equations:

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

Expanding this using the formula \( (p + q)^2 = p^2 + q^2 + 2pq \):

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

Now for the second equation:

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

Expanding this:

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

### Step 2: Combine the equations

Now we add both squared equations:

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

### Step 3: Simplify the equation

Combine like terms:

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

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

\[ x^2 + y^2 = a^2 \cdot 1 + b^2 \cdot 1 \]

Thus, we have:

\[ x^2 + y^2 = a^2 + b^2 \]

### Conclusion

Therefore, we conclude that:

\[ a^2 + b^2 = x^2 + y^2 \]

### Final Answer

The value of \( a^2 + b^2 \) is equal to \( x^2 + y^2 \).

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