If `x sintheta+ycostheta=2 & x sintheta-ycostheta=0` then which option is correct?

Options

A. `x^(2)+y^(2)=4`
B. `x^(2)+y^(2)=1`
C. `(1)/(x^(2))+(1)/(y^(2))=1`
D. `(1)/(x^(2))+(1)/(y^(2))=4`

Answer

To solve the equations \( x \sin \theta + y \cos \theta = 2 \) and \( x \sin \theta - y \cos \theta = 0 \), we can follow these steps:

### Step 1: Rewrite the second equation
From the second equation \( x \sin \theta - y \cos \theta = 0 \), we can express \( x \sin \theta \) in terms of \( y \cos \theta \):
\[ x \sin \theta = y \cos \theta \]

### Step 2: Substitute into the first equation
Now, substitute \( x \sin \theta \) from the second equation into the first equation:
\[ y \cos \theta + y \cos \theta = 2 \]
This simplifies to:
\[ 2y \cos \theta = 2 \]

### Step 3: Solve for \( \cos \theta \)
Dividing both sides by 2 gives:
\[ y \cos \theta = 1 \]
Thus, we can express \( \cos \theta \) as:
\[ \cos \theta = \frac{1}{y} \]

### Step 4: Substitute \( x \sin \theta \) back into the first equation
Now, we can find \( \sin \theta \) using the first equation. From \( x \sin \theta = y \cos \theta \), we have:
\[ \sin \theta = \frac{y \cos \theta}{x} \]
Substituting \( \cos \theta = \frac{1}{y} \):
\[ \sin \theta = \frac{y \cdot \frac{1}{y}}{x} = \frac{1}{x} \]

### Step 5: Use the Pythagorean identity
We know from trigonometric identities that:
\[ \cos^2 \theta + \sin^2 \theta = 1 \]
Substituting the values of \( \cos \theta \) and \( \sin \theta \):
\[ \left(\frac{1}{y}\right)^2 + \left(\frac{1}{x}\right)^2 = 1 \]
This simplifies to:
\[ \frac{1}{y^2} + \frac{1}{x^2} = 1 \]

### Conclusion
Thus, the correct relation is:
\[ \frac{1}{x^2} + \frac{1}{y^2} = 1 \]
This means the correct option is option 3.

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