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: Write down the equations
We start with the two equations:
1. \(a \sin \theta + b \cos \theta = x\) (Equation 1)
2. \(a \cos \theta - b \sin \theta = y\) (Equation 2)

### Step 2: Square both equations
Next, we square both equations and add them together:
\[ (a \sin \theta + b \cos \theta)^2 + (a \cos \theta - b \sin \theta)^2 = x^2 + y^2 \]

### Step 3: Expand the squared terms
Now, we expand both squared terms:
- For the first term:
\[ (a \sin \theta + b \cos \theta)^2 = a^2 \sin^2 \theta + 2ab \sin \theta \cos \theta + b^2 \cos^2 \theta \]
- For the second term:
\[ (a \cos \theta - b \sin \theta)^2 = a^2 \cos^2 \theta - 2ab \sin \theta \cos \theta + b^2 \sin^2 \theta \]

### Step 4: Combine the expanded terms
Adding these two expansions gives:
\[ (a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \sin \theta \cos \theta) + (a^2 \cos^2 \theta + b^2 \sin^2 \theta - 2ab \sin \theta \cos \theta) = x^2 + y^2 \]
The \(2ab \sin \theta \cos \theta\) and \(-2ab \sin \theta \cos \theta\) cancel out:
\[ a^2 \sin^2 \theta + b^2 \cos^2 \theta + a^2 \cos^2 \theta + b^2 \sin^2 \theta = x^2 + y^2 \]

### Step 5: Factor out common terms
Now, we can factor out \(a^2\) and \(b^2\):
\[ a^2 (\sin^2 \theta + \cos^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta) = x^2 + y^2 \]

### Step 6: Use the Pythagorean identity
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\[ a^2 (1) + b^2 (1) = x^2 + y^2 \]
This simplifies to:
\[ a^2 + b^2 = x^2 + y^2 \]

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