If `cottheta+tantheta=xand sectheta-costheta=y`, then

Options

A. `xsintheta.costheta=1`
B. `sin^2theta=ycostheta`
C. `(x^2y)^(1//3)+(xy^2)^(1//3)=1`
D. `(x^2y)^(2//3)-(xy^2)^(2//3)=1`

Explanation

To solve the problem given that \( \cot \theta + \tan \theta = x \) and \( \sec \theta - \cos \theta = y \), we will derive expressions for \( x \) and \( y \) in terms of \( \sin \theta \) and \( \cos \theta \), and then explore the relationships between them.

### Step-by-Step Solution:

1. **Express \( \cot \theta + \tan \theta \) in terms of sine and cosine:**
\[ \cot \theta = \frac{\cos \theta}{\sin \theta}, \quad \tan \theta = \frac{\sin \theta}{\cos \theta} \]
Therefore,
\[ \cot \theta + \tan \theta = \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} \]
Taking the common denominator:
\[ = \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} \]
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ = \frac{1}{\sin \theta \cos \theta} \]
Thus, we have:
\[ x = \frac{1}{\sin \theta \cos \theta} \]

2. **Express \( \sec \theta - \cos \theta \) in terms of sine and cosine:**
\[ \sec \theta = \frac{1}{\cos \theta} \]
Therefore,
\[ \sec \theta - \cos \theta = \frac{1}{\cos \theta} - \cos \theta \]
Taking the common denominator:
\[ = \frac{1 - \cos^2 \theta}{\cos \theta} \]
Using the identity \( 1 - \cos^2 \theta = \sin^2 \theta \):
\[ = \frac{\sin^2 \theta}{\cos \theta} \]
Thus, we have:
\[ y = \frac{\sin^2 \theta}{\cos \theta} \]

3. **Calculate \( x^2 y \):**
\[ x^2 = \left(\frac{1}{\sin \theta \cos \theta}\right)^2 = \frac{1}{\sin^2 \theta \cos^2 \theta} \]
Therefore,
\[ x^2 y = \frac{1}{\sin^2 \theta \cos^2 \theta} \cdot \frac{\sin^2 \theta}{\cos \theta} = \frac{1}{\cos^3 \theta} \]

4. **Calculate \( x y^2 \):**
\[ y^2 = \left(\frac{\sin^2 \theta}{\cos \theta}\right)^2 = \frac{\sin^4 \theta}{\cos^2 \theta} \]
Therefore,
\[ x y^2 = \frac{1}{\sin \theta \cos \theta} \cdot \frac{\sin^4 \theta}{\cos^2 \theta} = \frac{\sin^4 \theta}{\sin \theta \cos^3 \theta} = \frac{\sin^3 \theta}{\cos^3 \theta} \]

5. **Check the options:**
- For \( x^2 y \) we found \( \frac{1}{\cos^3 \theta} \).
- For \( x y^2 \) we found \( \frac{\sin^3 \theta}{\cos^3 \theta} \).

6. **Final relationships:**
- We can conclude that \( x^2 y = \frac{1}{\cos^3 \theta} \) and \( x y^2 = \frac{\sin^3 \theta}{\cos^3 \theta} \).
- Thus, we can verify the options provided in the original question.