If `x=a cos theta - b sin theta`, and `y=a sin theta+b cos theta`, what is the value of `(x^2+y^2)/(a^2+b^2)` ?

Answer

To solve the problem, we need to find the value of \(\frac{x^2 + y^2}{a^2 + b^2}\) given that \(x = a \cos \theta - b \sin \theta\) and \(y = a \sin \theta + b \cos \theta\).

### Step-by-Step Solution:

1. **Find \(x^2\)**:
\[ x = a \cos \theta - b \sin \theta \]
Squaring both sides:
\[ x^2 = (a \cos \theta - b \sin \theta)^2 \]
Expanding the square:
\[ x^2 = a^2 \cos^2 \theta - 2ab \cos \theta \sin \theta + b^2 \sin^2 \theta \]

2. **Find \(y^2\)**:
\[ y = a \sin \theta + b \cos \theta \]
Squaring both sides:
\[ y^2 = (a \sin \theta + b \cos \theta)^2 \]
Expanding the square:
\[ y^2 = a^2 \sin^2 \theta + 2ab \sin \theta \cos \theta + b^2 \cos^2 \theta \]

3. **Add \(x^2\) and \(y^2\)**:
\[ x^2 + y^2 = (a^2 \cos^2 \theta - 2ab \cos \theta \sin \theta + b^2 \sin^2 \theta) + (a^2 \sin^2 \theta + 2ab \sin \theta \cos \theta + b^2 \cos^2 \theta) \]
Combining like terms:
\[ x^2 + y^2 = a^2 (\cos^2 \theta + \sin^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta) \]
Using the identity \(\cos^2 \theta + \sin^2 \theta = 1\):
\[ x^2 + y^2 = a^2 \cdot 1 + b^2 \cdot 1 = a^2 + b^2 \]

4. **Calculate \(\frac{x^2 + y^2}{a^2 + b^2}\)**:
\[ \frac{x^2 + y^2}{a^2 + b^2} = \frac{a^2 + b^2}{a^2 + b^2} = 1 \]

### Final Answer:
\[ \frac{x^2 + y^2}{a^2 + b^2} = 1 \]