Eliminate `theta` from
`a sin theta + bcos theta = x "and" a cos theta -b sin theta = y`

Answer

To eliminate \( \theta \) from the equations \( a \sin \theta + b \cos \theta = x \) and \( a \cos \theta - b \sin \theta = y \), we can follow these steps:

### Step 1: Square both equations
First, we square both equations.

1. \( x = a \sin \theta + b \cos \theta \)
\[ x^2 = (a \sin \theta + b \cos \theta)^2 \]
Expanding this gives:
\[ x^2 = a^2 \sin^2 \theta + 2ab \sin \theta \cos \theta + b^2 \cos^2 \theta \]

2. \( y = a \cos \theta - b \sin \theta \)
\[ y^2 = (a \cos \theta - b \sin \theta)^2 \]
Expanding this gives:
\[ y^2 = a^2 \cos^2 \theta - 2ab \sin \theta \cos \theta + b^2 \sin^2 \theta \]

### Step 2: Add the squared equations
Now, we add \( x^2 \) and \( y^2 \):
\[ x^2 + y^2 = (a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \sin \theta \cos \theta) + (a^2 \cos^2 \theta + b^2 \sin^2 \theta - 2ab \sin \theta \cos \theta) \]

### Step 3: Simplify the expression
Combining like terms:
\[ x^2 + y^2 = a^2 (\sin^2 \theta + \cos^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta) \]
Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ x^2 + y^2 = a^2 \cdot 1 + b^2 \cdot 1 = a^2 + b^2 \]

### Final Result
Thus, we have eliminated \( \theta \) and the relationship between \( x \) and \( y \) is given by:
\[ x^2 + y^2 = a^2 + b^2 \]