If `x="cosec"theta-sinthetaandy=sectheta-costheta`, then the relationbetween x and y is

Options

A. `x^(2)+y^(2)+3=1`
B. `x^(2)y^(2)(x^(2)+y^(2)+3)=1`
C. `x^(2)(x^(2)+y^(2)-5)=1`
D. `y^(2)(x^(2)+y^(2)-5)=1`

Answer

To find the relation between \( x \) and \( y \) given \( x = \csc \theta - \sin \theta \) and \( y = \sec \theta - \cos \theta \), we will follow these steps:

### Step 1: Write down the expressions for \( x \) and \( y \).
Given:
\[ x = \csc \theta - \sin \theta \]
\[ y = \sec \theta - \cos \theta \]

### Step 2: Substitute \( \theta = 45^\circ \) into both expressions.
For \( \theta = 45^\circ \):
\[ \csc 45^\circ = \frac{1}{\sin 45^\circ} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2} \]
\[ \sin 45^\circ = \frac{1}{\sqrt{2}} \]
Thus,
\[ x = \sqrt{2} - \frac{1}{\sqrt{2}} = \sqrt{2} - \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]

Now for \( y \):
\[ \sec 45^\circ = \frac{1}{\cos 45^\circ} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2} \]
\[ \cos 45^\circ = \frac{1}{\sqrt{2}} \]
Thus,
\[ y = \sqrt{2} - \frac{1}{\sqrt{2}} = \sqrt{2} - \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Step 3: Calculate \( x^2 \) and \( y^2 \).
Now we have:
\[ x = \frac{1}{\sqrt{2}}, \quad y = \frac{1}{\sqrt{2}} \]
Calculating \( x^2 \) and \( y^2 \):
\[ x^2 = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]
\[ y^2 = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]

### Step 4: Find \( x^2 + y^2 + 3 \).
Now we calculate:
\[ x^2 + y^2 + 3 = \frac{1}{2} + \frac{1}{2} + 3 = 1 + 3 = 4 \]

### Step 5: Find \( x^2 y^2 \).
Calculating \( x^2 y^2 \):
\[ x^2 y^2 = \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) = \frac{1}{4} \]

### Step 6: Combine the results.
Now we can express the relation:
\[ x^2 y^2 + x^2 + y^2 + 3 = \frac{1}{4} + 1 + 3 = \frac{1}{4} + 4 = \frac{1}{4} + \frac{16}{4} = \frac{17}{4} \]

### Conclusion
The relation between \( x \) and \( y \) is:
\[ x^2 y^2 + x^2 + y^2 + 3 = 4 \]