If `x ="cosec"theta - sin theta` and `y=sectheta - costheta`, then the value of `x^(2)y^(2)(x^(2) + y^(2) + 3)` is:

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to find the value of \( x^2 y^2 (x^2 + y^2 + 3) \) given:

\[ x = \csc \theta - \sin \theta \]
\[ y = \sec \theta - \cos \theta \]

### Step 1: Simplifying \( x \)

We know that:
\[ \csc \theta = \frac{1}{\sin \theta} \]
Thus, we can rewrite \( x \):
\[ x = \frac{1}{\sin \theta} - \sin \theta \]
To combine these terms, we need a common denominator:
\[ x = \frac{1 - \sin^2 \theta}{\sin \theta} \]
Using the identity \( 1 - \sin^2 \theta = \cos^2 \theta \):
\[ x = \frac{\cos^2 \theta}{\sin \theta} = \cot \theta \cdot \cos \theta \]

### Step 2: Simplifying \( y \)

Similarly, for \( y \):
\[ \sec \theta = \frac{1}{\cos \theta} \]
So we can rewrite \( y \):
\[ y = \frac{1}{\cos \theta} - \cos \theta \]
Again, we need a common denominator:
\[ y = \frac{1 - \cos^2 \theta}{\cos \theta} \]
Using the identity \( 1 - \cos^2 \theta = \sin^2 \theta \):
\[ y = \frac{\sin^2 \theta}{\cos \theta} = \tan \theta \cdot \sin \theta \]

### Step 3: Finding \( x^2 \) and \( y^2 \)

Now we can find \( x^2 \) and \( y^2 \):
\[ x^2 = \left( \frac{\cos^2 \theta}{\sin \theta} \right)^2 = \frac{\cos^4 \theta}{\sin^2 \theta} \]
\[ y^2 = \left( \frac{\sin^2 \theta}{\cos \theta} \right)^2 = \frac{\sin^4 \theta}{\cos^2 \theta} \]

### Step 4: Finding \( x^2 + y^2 \)

Now we can find \( x^2 + y^2 \):
\[ x^2 + y^2 = \frac{\cos^4 \theta}{\sin^2 \theta} + \frac{\sin^4 \theta}{\cos^2 \theta} \]
To combine these, we need a common denominator:
\[ x^2 + y^2 = \frac{\cos^4 \theta \cdot \cos^2 \theta + \sin^4 \theta \cdot \sin^2 \theta}{\sin^2 \theta \cos^2 \theta} \]
This simplifies to:
\[ x^2 + y^2 = \frac{\cos^6 \theta + \sin^6 \theta}{\sin^2 \theta \cos^2 \theta} \]

### Step 5: Finding \( x^2 y^2 \)

Next, we find \( x^2 y^2 \):
\[ x^2 y^2 = \left( \frac{\cos^4 \theta}{\sin^2 \theta} \right) \left( \frac{\sin^4 \theta}{\cos^2 \theta} \right) = \frac{\cos^4 \theta \sin^4 \theta}{\sin^2 \theta \cos^2 \theta} = \cos^2 \theta \sin^2 \theta \]

### Step 6: Finding \( x^2 y^2 (x^2 + y^2 + 3) \)

Now we can substitute \( x^2 \), \( y^2 \), and \( 3 \) into the expression:
\[ x^2 + y^2 + 3 = \frac{\cos^6 \theta + \sin^6 \theta}{\sin^2 \theta \cos^2 \theta} + 3 \]
Thus, we have:
\[ x^2 y^2 (x^2 + y^2 + 3) = \cos^2 \theta \sin^2 \theta \left( \frac{\cos^6 \theta + \sin^6 \theta + 3 \sin^2 \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} \right) \]
This simplifies to:
\[ \cos^6 \theta + \sin^6 \theta + 3 \sin^2 \theta \cos^2 \theta \]

### Final Step: Conclusion

The final expression is:
\[ \cos^6 \theta + \sin^6 \theta + 3 \sin^2 \theta \cos^2 \theta \]