If `x=sectheta-tantheta`, and `y=cosectheta+cottheta`, then

Options

A. `xy+1=x-y`
B. `xy+1=x-2y`
C. `xy+1=x+y`
D. `xy+1=y-x`

Answer

To solve the problem, we need to analyze the expressions given for \( x \) and \( y \), and then check which option is correct by substituting the values.

### Step 1: Define the expressions
We have:
\[ x = \sec \theta - \tan \theta \]
\[ y = \csc \theta + \cot \theta \]

### Step 2: Substitute \( \theta = 45^\circ \)
We will substitute \( \theta = 45^\circ \) into the expressions for \( x \) and \( y \).

Calculating \( x \):
\[ \sec 45^\circ = \frac{1}{\cos 45^\circ} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2} \]
\[ \tan 45^\circ = \frac{\sin 45^\circ}{\cos 45^\circ} = 1 \]
Thus,
\[ x = \sec 45^\circ - \tan 45^\circ = \sqrt{2} - 1 \]

Calculating \( y \):
\[ \csc 45^\circ = \frac{1}{\sin 45^\circ} = \sqrt{2} \]
\[ \cot 45^\circ = \frac{\cos 45^\circ}{\sin 45^\circ} = 1 \]
Thus,
\[ y = \csc 45^\circ + \cot 45^\circ = \sqrt{2} + 1 \]

### Step 3: Calculate \( xy + 1 \)
Now we need to calculate \( xy + 1 \):
\[ xy = (\sqrt{2} - 1)(\sqrt{2} + 1) \]
Using the difference of squares:
\[ xy = (\sqrt{2})^2 - (1)^2 = 2 - 1 = 1 \]
Thus,
\[ xy + 1 = 1 + 1 = 2 \]

### Step 4: Check the options
Now we need to check each option to see which one equals \( 2 \).

**Option 1:** \( xy + 1 = x - y \)
\[ x - y = (\sqrt{2} - 1) - (\sqrt{2} + 1) = \sqrt{2} - 1 - \sqrt{2} - 1 = -2 \quad \text{(not equal to 2)} \]

**Option 2:** \( xy + 1 = x - 2y \)
\[ x - 2y = (\sqrt{2} - 1) - 2(\sqrt{2} + 1) = \sqrt{2} - 1 - 2\sqrt{2} - 2 = -\sqrt{2} - 3 \quad \text{(not equal to 2)} \]

**Option 3:** \( xy + 1 = x + y \)
\[ x + y = (\sqrt{2} - 1) + (\sqrt{2} + 1) = 2\sqrt{2} \quad \text{(not equal to 2)} \]

**Option 4:** \( xy + 1 = y - x \)
\[ y - x = (\sqrt{2} + 1) - (\sqrt{2} - 1) = \sqrt{2} + 1 - \sqrt{2} + 1 = 2 \quad \text{(equal to 2)} \]

### Conclusion
The correct option is:
\[ \text{Option 4: } xy + 1 = y - x \]