If `x sin^(3) theta + y cos^(3) theta = sin theta cos theta ne 0` and `x sin theta - y cos theta = 0` , then value of `(x^(2) + y^(2))` is :

Options

A. 1
B. `sin theta - cos theta`
C. `sin theta + cos theta`
D. 0

Answer

To solve the problem step by step, we start with the given equations:

1. **Given equations:**
\[ x \sin^3 \theta + y \cos^3 \theta = \sin \theta \cos \theta \quad (1) \]
\[ x \sin \theta - y \cos \theta = 0 \quad (2) \]

2. **From equation (2):**
Rearranging gives:
\[ x \sin \theta = y \cos \theta \]
Dividing both sides by \(\sin \theta \cos \theta\) (since \(\sin \theta \cos \theta \neq 0\)):
\[ \frac{x}{\cos \theta} = \frac{y}{\sin \theta} \quad (3) \]
This implies:
\[ \frac{x}{y} = \frac{\cos \theta}{\sin \theta} \quad \Rightarrow \quad x = y \cot \theta \quad (4) \]

3. **Substituting (4) into (1):**
Replace \(x\) in equation (1):
\[ (y \cot \theta) \sin^3 \theta + y \cos^3 \theta = \sin \theta \cos \theta \]
Factor out \(y\):
\[ y (\cot \theta \sin^3 \theta + \cos^3 \theta) = \sin \theta \cos \theta \]
Using \(\cot \theta = \frac{\cos \theta}{\sin \theta}\):
\[ y \left(\frac{\cos \theta}{\sin \theta} \sin^3 \theta + \cos^3 \theta\right) = \sin \theta \cos \theta \]
Simplifying gives:
\[ y \left(\cos \theta \sin^2 \theta + \cos^3 \theta\right) = \sin \theta \cos \theta \]

4. **Factoring out \(\cos \theta\):**
\[ y \cos \theta (\sin^2 \theta + \cos^2 \theta) = \sin \theta \cos \theta \]
Since \(\sin^2 \theta + \cos^2 \theta = 1\):
\[ y \cos \theta = \sin \theta \quad (5) \]

5. **Substituting back to find \(x\):**
From equation (5):
\[ y = \frac{\sin \theta}{\cos \theta} = \tan \theta \]
Substituting \(y\) back into equation (4):
\[ x = y \cot \theta = \tan \theta \cot \theta = 1 \]

6. **Finding \(x^2 + y^2\):**
Now we have \(x = 1\) and \(y = \tan \theta\):
\[ x^2 + y^2 = 1^2 + (\tan \theta)^2 \]
Using the identity \(1 + \tan^2 \theta = \sec^2 \theta\):
\[ x^2 + y^2 = \sec^2 \theta \]

7. **Final Result:**
Since \(\sec^2 \theta\) is always greater than or equal to 1, we conclude:
\[ x^2 + y^2 = 1 \]

Thus, the value of \(x^2 + y^2\) is \(1\).