Find `( sin theta - sin 3 theta )/( sin^(2) theta - cos^(2) theta)`

Options

A. `2 cos theta `
B. `-2 sin theta `
C. `2 sin theta `
D. `-2 cos theta `

Answer

To find the value of \( \frac{\sin \theta - \sin 3\theta}{\sin^2 \theta - \cos^2 \theta} \), we can follow these steps:

### Step 1: Use the identity for \(\sin 3\theta\)
We know that:
\[ \sin 3\theta = 3\sin\theta - 4\sin^3\theta \]
Substituting this into our expression gives:
\[ \sin \theta - \sin 3\theta = \sin \theta - (3\sin \theta - 4\sin^3 \theta) = \sin \theta - 3\sin \theta + 4\sin^3 \theta \]
This simplifies to:
\[ -2\sin \theta + 4\sin^3 \theta \]

### Step 2: Rewrite the denominator
The denominator can be rewritten using the identity:
\[ \sin^2 \theta - \cos^2 \theta = -(\cos^2 \theta - \sin^2 \theta) = -\cos 2\theta \]
Thus, we have:
\[ \sin^2 \theta - \cos^2 \theta = -\cos 2\theta \]

### Step 3: Substitute back into the expression
Now substituting both the numerator and the denominator back into the expression gives:
\[ \frac{-2\sin \theta + 4\sin^3 \theta}{-\cos 2\theta} \]
This simplifies to:
\[ \frac{2\sin \theta - 4\sin^3 \theta}{\cos 2\theta} \]

### Step 4: Factor the numerator
We can factor out \(2\sin \theta\) from the numerator:
\[ 2\sin \theta(1 - 2\sin^2 \theta) \]
Thus, we have:
\[ \frac{2\sin \theta(1 - 2\sin^2 \theta)}{\cos 2\theta} \]

### Step 5: Use the identity for \(1 - 2\sin^2 \theta\)
We know that:
\[ 1 - 2\sin^2 \theta = \cos 2\theta \]
So we can substitute this into our expression:
\[ \frac{2\sin \theta \cos 2\theta}{\cos 2\theta} \]

### Step 6: Cancel out \(\cos 2\theta\)
Assuming \(\cos 2\theta \neq 0\), we can cancel \(\cos 2\theta\) from the numerator and denominator:
\[ 2\sin \theta \]

### Final Answer
Thus, the value of the expression is:
\[ \boxed{2\sin \theta} \]