If `x=sec theta-tan theta `and `y=cosectheta+cot theta`,then Find
`y-x-xy-1=`

Answer

To solve the problem, we need to find the expression \( y - x - xy - 1 \) given \( x = \sec \theta - \tan \theta \) and \( y = \csc \theta + \cot \theta \).

### Step 1: Substitute the values of \( x \) and \( y \)

We have:
\[ x = \sec \theta - \tan \theta \]
\[ y = \csc \theta + \cot \theta \]

### Step 2: Rewrite \( x \) and \( y \) in terms of sine and cosine

Using the definitions of secant, tangent, cosecant, and cotangent:
\[ \sec \theta = \frac{1}{\cos \theta}, \quad \tan \theta = \frac{\sin \theta}{\cos \theta} \]
\[ \csc \theta = \frac{1}{\sin \theta}, \quad \cot \theta = \frac{\cos \theta}{\sin \theta} \]

So we can rewrite \( x \) and \( y \):
\[ x = \frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta} = \frac{1 - \sin \theta}{\cos \theta} \]
\[ y = \frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta} = \frac{1 + \cos \theta}{\sin \theta} \]

### Step 3: Calculate \( xy \)

Now, we will calculate \( xy \):
\[ xy = \left( \frac{1 - \sin \theta}{\cos \theta} \right) \left( \frac{1 + \cos \theta}{\sin \theta} \right) \]
\[ = \frac{(1 - \sin \theta)(1 + \cos \theta)}{\sin \theta \cos \theta} \]

Expanding the numerator:
\[ = \frac{1 + \cos \theta - \sin \theta - \sin \theta \cos \theta}{\sin \theta \cos \theta} \]

### Step 4: Combine \( y - x - xy - 1 \)

Now we need to compute \( y - x - xy - 1 \):
\[ y - x = \frac{1 + \cos \theta}{\sin \theta} - \frac{1 - \sin \theta}{\cos \theta} \]
To combine these fractions, we will find a common denominator:
\[ = \frac{(1 + \cos \theta) \cos \theta - (1 - \sin \theta) \sin \theta}{\sin \theta \cos \theta} \]
Expanding the numerator:
\[ = \frac{\cos \theta + \cos^2 \theta - \sin \theta + \sin^2 \theta}{\sin \theta \cos \theta} \]
Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ = \frac{1 + \cos \theta - \sin \theta}{\sin \theta \cos \theta} \]

Now, we can substitute this back into our expression:
\[ y - x - xy - 1 = \left( \frac{1 + \cos \theta - \sin \theta}{\sin \theta \cos \theta} \right) - \left( \frac{1 + \cos \theta - \sin \theta}{\sin \theta \cos \theta} + 1 \right) \]
This simplifies to:
\[ = 0 \]

### Final Answer
Thus, we conclude that:
\[ y - x - xy - 1 = 0 \]