If `x=asintheta-bcostheta,y=acostheta+bsintheta` , then which of the following is true ?

Options

A. `(x^(2))/(y^(2))+(a^(2))/(b^(2))=1`
B. `x^(2)+y^(2)=a^(2)-b^(2)`
C. `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`
D. `x^(2)+y^(2)=a^(2)+b^(2)`

Answer

To solve the problem, we need to find the relationship between \( x \) and \( y \) given the equations:

\[ x = a \sin \theta - b \cos \theta \]
\[ y = a \cos \theta + b \sin \theta \]

### Step 1: Calculate \( x^2 \)

First, we will square \( x \):

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

Expanding this using the formula \( (A - B)^2 = A^2 - 2AB + B^2 \):

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

This simplifies to:

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

### Step 2: Calculate \( y^2 \)

Now, we will square \( y \):

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

Expanding this using the formula \( (A + B)^2 = A^2 + 2AB + B^2 \):

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

This simplifies to:

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

### Step 3: Add \( x^2 \) and \( y^2 \)

Now, we will add \( x^2 \) and \( y^2 \):

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

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

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

### Conclusion

Thus, we have shown that:

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

This means that the relationship between \( x \) and \( y \) is given by the equation \( x^2 + y^2 = a^2 + b^2 \).