The eliminant of `theta` from x cos `theta` - y sin `theta` = 2 , x sin `theta` + y cos `theta` = 4 will give

Options

A. `x^2 + y^2 = 20`
B. `3x^2 + y^2 = 20`
C. `x^2 - y^2 = 20`
D. `x^2-y^2 =20`

Answer

To find the eliminant of \( \theta \) from the equations \( x \cos \theta - y \sin \theta = 2 \) and \( x \sin \theta + y \cos \theta = 4 \), we can follow these steps:

### Step 1: Square the first equation
We start with the first equation:
\[ x \cos \theta - y \sin \theta = 2 \]
Squaring both sides gives:
\[ (x \cos \theta - y \sin \theta)^2 = 2^2 \]
Expanding the left side:
\[ x^2 \cos^2 \theta - 2xy \cos \theta \sin \theta + y^2 \sin^2 \theta = 4 \]

### Step 2: Square the second equation
Now take the second equation:
\[ x \sin \theta + y \cos \theta = 4 \]
Squaring both sides gives:
\[ (x \sin \theta + y \cos \theta)^2 = 4^2 \]
Expanding the left side:
\[ x^2 \sin^2 \theta + 2xy \sin \theta \cos \theta + y^2 \cos^2 \theta = 16 \]

### Step 3: Add both squared equations
Now we add the two squared equations:
\[ (x^2 \cos^2 \theta - 2xy \cos \theta \sin \theta + y^2 \sin^2 \theta) + (x^2 \sin^2 \theta + 2xy \sin \theta \cos \theta + y^2 \cos^2 \theta) = 4 + 16 \]
This simplifies to:
\[ x^2 \cos^2 \theta + y^2 \cos^2 \theta + x^2 \sin^2 \theta + y^2 \sin^2 \theta = 20 \]

### Step 4: Use the Pythagorean identity
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ x^2 (\sin^2 \theta + \cos^2 \theta) + y^2 (\sin^2 \theta + \cos^2 \theta) = 20 \]
This simplifies to:
\[ x^2 + y^2 = 20 \]

### Conclusion
The eliminant of \( \theta \) from the given equations is:
\[ x^2 + y^2 = 20 \]