To solve the given problem, let's first understand the expressions and relationships between \(\sec\theta\), \(\tan\theta\), \text{cosec}\theta\), and \(\cot\theta\\)

1. We are given:
   • \(x = \sec\theta - \tan\theta\)
   • \(y = \text{cosec}\theta + \cot\theta\)
2. We need to verify which of the given equations are satisfied by these expressions. Let us first work with the expressions for \(x\) and \(y\).
3. Rewriting \(\sec\theta\) and \(\tan\theta\) using basic trigonometric identities:
   • \(\sec\theta = \frac{1}{\cos\theta}\)
   • \(\tan\theta = \frac{\sin\theta}{\cos\theta}\)
4. Substituting in \(x = \sec\theta - \tan\theta\), we get: \(x = \frac{1}{\cos\theta} - \frac{\sin\theta}{\cos\theta} = \frac{1 - \sin\theta}{\cos\theta}\)
5. Similarly, rewrite \(\text{cosec}\theta\) and \(\cot\theta\):
   • \(\text{cosec}\theta = \frac{1}{\sin\theta}\)
   • \(\cot\theta = \frac{\cos\theta}{\sin\theta}\)
6. Substituting in \(y = \text{cosec}\theta + \cot\theta\), we get: \(y = \frac{1}{\sin\theta} + \frac{\cos\theta}{\sin\theta} = \frac{1 + \cos\theta}{\sin\theta}\)
7. We now evaluate \(x - y + xy + 1\):
   • Substitute the expressions for \(x\) and \(y\): \(x - y = \frac{1 - \sin\theta}{\cos\theta} - \frac{1 + \cos\theta}{\sin\theta}\)
   • Find the LCM of cosθ and sinθ: \(x - y = \frac{(1 - \sin\theta)\sin\theta - (1 + \cos\theta)\cos\theta}{\cos\theta \cdot \sin\theta}\)
   • Calculate \(xy\): \(xy = \left(\frac{1 - \sin\theta}{\cos\theta}\right) \left(\frac{1 + \cos\theta}{\sin\theta}\right) = \frac{(1 - \sin\theta)(1 + \cos\theta)}{\cos\theta \cdot \sin\theta}\)
   • Finally, evaluate: \(x - y + xy + 1 = \frac{(1 - \sin\theta)\sin\theta - (1 + \cos\theta)\cos\theta + (1 - \sin\theta)(1 + \cos\theta) + \cos\theta \cdot \sin\theta}{\cos\theta \cdot \sin\theta}\)
8. The result will simplify to: \(x - y + xy + 1 = 0\)

Thus, the correct answer is: \(x - y + xy + 1 = 0\)