The expression `(sin 8thetacos theta-sin6 thetacos 3 theta)/(cos 2 thetacos theta-sin 3 thetasin 4 theta)` is equals:-

Options

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

Answer

To solve the expression

\[ \frac{\sin 8\theta \cos \theta - \sin 6\theta \cos 3\theta}{\cos 2\theta \cos \theta - \sin 3\theta \sin 4\theta} \]

we can follow these steps:

### Step 1: Rewrite the expression
Let \( x = \frac{\sin 8\theta \cos \theta - \sin 6\theta \cos 3\theta}{\cos 2\theta \cos \theta - \sin 3\theta \sin 4\theta} \).

### Step 2: Multiply numerator and denominator by 2
To simplify, we multiply both the numerator and denominator by 2:

\[ x = \frac{2(\sin 8\theta \cos \theta - \sin 6\theta \cos 3\theta)}{2(\cos 2\theta \cos \theta - \sin 3\theta \sin 4\theta)} \]

### Step 3: Apply the sine and cosine identities
Using the identities:
- \( 2 \sin A \cos B = \sin(A + B) + \sin(A - B) \)
- \( 2 \cos A \cos B = \cos(A + B) + \cos(A - B) \)
- \( 2 \sin A \sin B = \cos(A - B) - \cos(A + B) \)

We can simplify the numerator and denominator.

#### Numerator:
1. For \( \sin 8\theta \cos \theta \):
\[ 2 \sin 8\theta \cos \theta = \sin(8\theta + \theta) + \sin(8\theta - \theta) = \sin 9\theta + \sin 7\theta \]
2. For \( \sin 6\theta \cos 3\theta \):
\[ 2 \sin 6\theta \cos 3\theta = \sin(6\theta + 3\theta) + \sin(6\theta - 3\theta) = \sin 9\theta + \sin 3\theta \]
3. Thus, the numerator becomes:
\[ \sin 9\theta + \sin 7\theta - (\sin 9\theta + \sin 3\theta) = \sin 7\theta - \sin 3\theta \]

#### Denominator:
1. For \( \cos 2\theta \cos \theta \):
\[ 2 \cos 2\theta \cos \theta = \cos(2\theta + \theta) + \cos(2\theta - \theta) = \cos 3\theta + \cos \theta \]
2. For \( \sin 3\theta \sin 4\theta \):
\[ 2 \sin 3\theta \sin 4\theta = \cos(3\theta - 4\theta) - \cos(3\theta + 4\theta) = \cos(-\theta) - \cos(7\theta) = \cos \theta - \cos 7\theta \]
3. Thus, the denominator becomes:
\[ \cos 3\theta + \cos \theta - (\cos \theta - \cos 7\theta) = \cos 3\theta + \cos 7\theta \]

### Step 4: Simplify the expression
Now we have:

\[ x = \frac{\sin 7\theta - \sin 3\theta}{\cos 3\theta + \cos 7\theta} \]

### Step 5: Apply sine and cosine difference formulas
Using the identities:
- \( \sin A - \sin B = 2 \cos\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) \)
- \( \cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \)

We can rewrite the numerator and denominator:

1. Numerator:
\[ \sin 7\theta - \sin 3\theta = 2 \cos\left(\frac{7\theta + 3\theta}{2}\right) \sin\left(\frac{7\theta - 3\theta}{2}\right) = 2 \cos(5\theta) \sin(2\theta) \]

2. Denominator:
\[ \cos 3\theta + \cos 7\theta = 2 \cos\left(\frac{3\theta + 7\theta}{2}\right) \cos\left(\frac{7\theta - 3\theta}{2}\right) = 2 \cos(5\theta) \cos(2\theta) \]

### Step 6: Final simplification
Now substituting these back into the expression:

\[ x = \frac{2 \cos(5\theta) \sin(2\theta)}{2 \cos(5\theta) \cos(2\theta)} \]

The \( 2 \cos(5\theta) \) terms cancel out (assuming \( \cos(5\theta) \neq 0 \)):

\[ x = \frac{\sin(2\theta)}{\cos(2\theta)} = \tan(2\theta) \]

### Conclusion
Thus, the expression simplifies to:

\[ \boxed{\tan(2\theta)} \]